转自kuangbin的AC自动机（赛前最后一博）

有了KMP和Trie的基础，就可以学习神奇的AC自动机了。AC自动机其实就是在Trie树上实现KMP，可以完成多模式串的匹配。

AC自动机其实就是创建了一个状态的转移图，思想很重要。

http://blog.csdn.net/niushuai666/article/details/7002823

http://hi.baidu.com/nialv7/item/ce1ce015d44a6ba7feded52d

AC自动机专题训练链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=25605#overview 这里我提交的代码是公开的，可以看到

题目来源：http://www.notonlysuccess.com/index.php/aho-corasick-automaton/

写AC自动机的代码风格是向昀神学的，好简洁，写起来好棒的感觉。

1、HDU 2222 Keywords Search 最基本的入门题了

就是求目标串中出现了几个模式串。

很基础了。使用一个int型的end数组记录，查询一次。

//======================
// HDU 2222
// 求目标串中出现了几个模式串
//====================
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;

struct Trie
{
    int next[500010][26],fail[500010],end[500010];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 26;i++)
            next[L][i] = -1;
        end[L++] = 0;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][buf[i]-'a'] == -1)
                next[now][buf[i]-'a'] = newnode();
            now = next[now][buf[i]-'a'];
        }
        end[now]++;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 26;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while( !Q.empty() )
        {
            int now = Q.front();
            Q.pop();
            for(int i = 0;i < 26;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]]=next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    int query(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        int res = 0;
        for(int i = 0;i < len;i++)
        {
            now = next[now][buf[i]-'a'];
            int temp = now;
            while( temp != root )
            {
                res += end[temp];
                end[temp] = 0;
                temp = fail[temp];
            }
        }
        return res;
    }
    void debug()
    {
        for(int i = 0;i < L;i++)
        {
            printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
            for(int j = 0;j < 26;j++)
                printf("%2d",next[i][j]);
            printf("]
");
        }
    }
};
char buf[1000010];
Trie ac;
int main()
{
    int T;
    int n;
    scanf("%d",&T);
    while( T-- )
    {
        scanf("%d",&n);
        ac.init();
        for(int i = 0;i < n;i++)
        {
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        scanf("%s",buf);
        printf("%d
",ac.query(buf));
    }
    return 0;
}

2、HDU 2896 病毒侵袭

这题和上题差不多，要输出出现模式串的id,用end记录id就可以了。还有trie树的分支是128的

题解here

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;

struct Trie
{
    int next[210*500][128],fail[210*500],end[210*500];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 128;i++)
            next[L][i] = -1;
        end[L++] = -1;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char s[],int id)
    {
        int len = strlen(s);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][s[i]] == -1)
                next[now][s[i]] = newnode();
            now=next[now][s[i]];
        }
        end[now]=id;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 128;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            for(int i = 0;i < 128;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    bool used[510];
    bool query(char buf[],int n,int id)
    {
        int len = strlen(buf);
        int now = root;
        memset(used,false,sizeof(used));
        bool flag = false;
        for(int i = 0;i < len;i++)
        {
            now = next[now][buf[i]];
            int temp = now;
            while(temp != root)
            {
                if(end[temp] != -1)
                {
                    used[end[temp]] = true;
                    flag = true;
                }
                temp = fail[temp];
            }
        }
        if(!flag)return false;
        printf("web %d:",id);
        for(int i = 1;i <= n;i++)
            if(used[i])
                printf(" %d",i);
        printf("
");
        return true;
    }
};
char buf[10010];
Trie ac;
int main()
{
    int n,m;
    while(scanf("%d",&n) != EOF)
    {
        ac.init();
        for(int i = 1;i <= n;i++)
        {
            scanf("%s",buf);
            ac.insert(buf,i);
        }
        ac.build();
        int ans = 0;
        scanf("%d",&m);
        for(int i = 1;i <= m;i++)
        {
            scanf("%s",buf);
            if(ac.query(buf,n,i))
                ans++;
        }
        printf("total: %d
",ans);
    }
    return 0;
}

View Code

3、HDU 3065 病毒侵袭持续中

这题的变化也不大，就是需要输出每个模式串出现的次数，查询的时候使用一个数组进行记录就可以了

  1 //============================================================================
  2 // Name        : HDU.cpp
  3 // Author      : 
  4 // Version     :
  5 // Copyright   : Your copyright notice
  6 // Description : Hello World in C++, Ansi-style
  7 //============================================================================
  8 
  9 #include <iostream>
 10 #include <stdio.h>
 11 #include <string.h>
 12 #include <algorithm>
 13 #include <queue>
 14 using namespace std;
 15 
 16 char str[1010][100];
 17 struct Trie
 18 {
 19     int next[1010*50][128],fail[1010*50],end[1010*50];
 20     int root,L;
 21     int newnode()
 22     {
 23         for(int i = 0;i < 128;i++)
 24             next[L][i] = -1;
 25         end[L++] = -1;
 26         return L-1;
 27     }
 28     void init()
 29     {
 30         L = 0;
 31         root = newnode();
 32     }
 33     void insert(char s[],int id)
 34     {
 35         int len = strlen(s);
 36         int now = root;
 37         for(int i = 0;i < len;i++)
 38         {
 39             if(next[now][s[i]] == -1)
 40                 next[now][s[i]] = newnode();
 41             now = next[now][s[i]];
 42         }
 43         end[now] = id;
 44     }
 45     void build()
 46     {
 47         queue<int>Q;
 48         fail[root] = root;
 49         for(int i = 0;i < 128;i++)
 50             if(next[root][i] == -1)
 51                 next[root][i] = root;
 52             else
 53             {
 54                 fail[next[root][i]] = root;
 55                 Q.push(next[root][i]);
 56             }
 57         while(!Q.empty())
 58         {
 59             int now = Q.front();
 60             Q.pop();
 61             for(int i = 0;i < 128;i++)
 62                 if(next[now][i] == -1)
 63                     next[now][i]=next[fail[now]][i];
 64                 else
 65                 {
 66                     fail[next[now][i]]=next[fail[now]][i];
 67                     Q.push(next[now][i]);
 68                 }
 69         }
 70     }
 71     int num[1010];
 72     void query(char buf[],int n)
 73     {
 74         for(int i = 0;i < n;i++)
 75             num[i] = 0;
 76         int len=strlen(buf);
 77         int now=root;
 78         for(int i=0;i<len;i++)
 79         {
 80             now=next[now][buf[i]];
 81             int temp = now;
 82             while( temp != root )
 83             {
 84                 if(end[temp] != -1)
 85                     num[end[temp]]++;
 86                 temp = fail[temp];
 87             }
 88         }
 89         for(int i = 0;i < n;i++)
 90             if(num[i] > 0)
 91                 printf("%s: %d
",str[i],num[i]);
 92     }
 93 
 94 };
 95 
 96 char buf[2000010];
 97 Trie ac;
 98 void debug()
 99 {
100     for (int i = 0; i < ac.L; i++)
101     {
102         printf("id = %3d ,fail = %3d ,end = %3d, chi = [",i,ac.fail[i],ac.end[i]);
103         for (int j = 0; j < 128; j++)
104             printf("%2d ",ac.next[i][j]);
105         printf("]
");
106     }
107 }
108 int main()
109 {
110 //    freopen("in.txt","r",stdin);
111 //    freopen("out.txt","w",stdout);
112     int n;
113     while(scanf("%d",&n) == 1)
114     {
115         ac.init();
116         for(int i = 0;i < n;i++)
117         {
118             scanf("%s",str[i]);
119             ac.insert(str[i],i);
120         }
121         ac.build();
122         scanf("%s",buf);
123         ac.query(buf,n);
124     }
125     return 0;
126 }

View Code

4、ZOJ 3430 Detect the Virus

主要是解码过程，解码以后就是模板题了。

求出现的模式串的种类数

分支需要256个

  1 //============================================================================
  2 // Name        : ZOJ.cpp
  3 // Author      : 
  4 // Version     :
  5 // Copyright   : Your copyright notice
  6 // Description : Hello World in C++, Ansi-style
  7 //============================================================================
  8 
  9 #include <iostream>
 10 #include <stdio.h>
 11 #include <string.h>
 12 #include <algorithm>
 13 #include <queue>
 14 using namespace std;
 15 
 16 struct Trie
 17 {
 18     int next[520*64][256],fail[520*64],end[520*64];
 19     int root,L;
 20     int newnode()
 21     {
 22         for(int i = 0;i < 256;i++)
 23             next[L][i] = -1;
 24         end[L++] = -1;
 25         return L-1;
 26     }
 27     void init()
 28     {
 29         L = 0;
 30         root = newnode();
 31     }
 32     void insert(unsigned char buf[],int len,int id)
 33     {
 34         int now = root;
 35         for(int i = 0;i < len;i++)
 36         {
 37             if(next[now][buf[i]] == -1)
 38                 next[now][buf[i]] = newnode();
 39             now = next[now][buf[i]];
 40         }
 41         end[now] = id;
 42     }
 43     void build()
 44     {
 45         queue<int>Q;
 46         fail[root] = root;
 47         for(int i = 0;i < 256;i++)
 48             if(next[root][i] == -1)
 49                 next[root][i] = root;
 50             else
 51             {
 52                 fail[next[root][i]]=root;
 53                 Q.push(next[root][i]);
 54             }
 55         while(!Q.empty())
 56         {
 57             int now = Q.front();
 58             Q.pop();
 59             for(int i = 0;i < 256;i++)
 60                 if(next[now][i] == -1)
 61                     next[now][i] = next[fail[now]][i];
 62                 else
 63                 {
 64                     fail[next[now][i]] = next[fail[now]][i];
 65                     Q.push(next[now][i]);
 66                 }
 67         }
 68     }
 69     bool used[520];
 70     int query(unsigned char buf[],int len,int n)
 71     {
 72         memset(used,false,sizeof(used));
 73         int now = root;
 74         for(int i = 0;i < len;i++)
 75         {
 76             now = next[now][buf[i]];
 77             int temp = now;
 78             while( temp!=root )
 79             {
 80                 if(end[temp] != -1)
 81                     used[end[temp]]=true;
 82                 temp = fail[temp];
 83             }
 84         }
 85         int res = 0;
 86         for(int i = 0;i < n;i++)
 87             if(used[i])
 88                 res++;
 89         return res;
 90     }
 91 };
 92 
 93 unsigned char buf[2050];
 94 int tot;
 95 char str[4000];
 96 unsigned char s[4000];
 97 unsigned char Get(char ch)
 98 {
 99     if( ch>='A'&&ch<='Z' )return ch-'A';
100     if( ch>='a'&&ch<='z' )return ch-'a'+26;
101     if( ch>='0'&&ch<='9' )return ch-'0'+52;
102     if( ch=='+' )return 62;
103     else return 63;
104 }
105 void change(unsigned char str[],int len)
106 {
107     int t=0;
108     for(int i=0;i<len;i+=4)
109     {
110         buf[t++]=((str[i]<<2)|(str[i+1]>>4));
111         if(i+2 < len)
112             buf[t++]=( (str[i+1]<<4)|(str[i+2]>>2) );
113         if(i+3 < len)
114             buf[t++]= ( (str[i+2]<<6)|str[i+3] );
115     }
116     tot=t;
117 }
118 Trie ac;
119 int main()
120 {
121 //    freopen("in.txt","r",stdin);
122 //    freopen("out.txt","w",stdout);
123     int n,m;
124     while(scanf("%d",&n) == 1)
125     {
126         ac.init();
127         for(int i = 0;i < n;i++)
128         {
129             scanf("%s",str);
130             int len = strlen(str);
131             while(str[len-1]=='=')len--;
132             for(int j = 0;j < len;j++)
133             {
134                 s[j] = Get(str[j]);
135             }
136             change(s,len);
137             ac.insert(buf,tot,i);
138         }
139         ac.build();
140         scanf("%d",&m);
141         while(m--)
142         {
143             scanf("%s",str);
144             int len=strlen(str);
145             while(str[len-1]=='=')len--;
146             for(int j = 0;j < len;j++)
147                 s[j] = Get(str[j]);
148             change(s,len);
149             printf("%d
",ac.query(buf,tot,n));
150         }
151         printf("
");
152     }
153     return 0;
154 }

View Code

5、POJ 2778 DNA Sequence

AC自动机+矩阵加速

这个时候AC自动机的一种状态转移图的思路就很透彻了。

AC自动机就是可以确定状态的转移。

  1 //============================================================================
  2 // Name        : POJ.cpp
  3 // Author      : 
  4 // Version     :
  5 // Copyright   : Your copyright notice
  6 // Description : Hello World in C++, Ansi-style
  7 //============================================================================
  8 
  9 #include <iostream>
 10 #include <stdio.h>
 11 #include <algorithm>
 12 #include <string.h>
 13 #include <queue>
 14 using namespace std;
 15 
 16 const int MOD=100000;
 17 struct Matrix
 18 {
 19     int mat[110][110],n;
 20     Matrix(){}
 21     Matrix(int _n)
 22     {
 23         n = _n;
 24         for(int i=0;i<n;i++)
 25             for(int j=0;j<n;j++)
 26                 mat[i][j]=0;
 27     }
 28     Matrix operator *(const Matrix &b)const
 29     {
 30         Matrix ret=Matrix(n);
 31         for(int i=0;i<n;i++)
 32             for(int j=0;j<n;j++)
 33                 for(int k=0;k<n;k++)
 34                 {
 35                     int tmp=(long long)mat[i][k]*b.mat[k][j]%MOD;
 36                     ret.mat[i][j]=(ret.mat[i][j]+tmp)%MOD;
 37                 }
 38         return ret;
 39     }
 40 };
 41 struct Trie
 42 {
 43     int next[110][4],fail[110];
 44     bool end[110];
 45     int root,L;
 46     int newnode()
 47     {
 48         for(int i=0;i<4;i++)
 49             next[L][i]=-1;
 50         end[L++]=false;
 51         return L-1;
 52     }
 53     void init()
 54     {
 55         L=0;
 56         root=newnode();
 57     }
 58     int getch(char ch)
 59     {
 60         switch(ch)
 61         {
 62         case 'A':
 63             return 0;
 64             break;
 65         case 'C':
 66             return 1;
 67             break;
 68         case 'G':
 69             return 2;
 70             break;
 71         case 'T':
 72             return 3;
 73             break;
 74         }
 75     }
 76     void insert(char s[])
 77     {
 78         int len=strlen(s);
 79         int now=root;
 80         for(int i = 0;i < len;i++)
 81         {
 82             if(next[now][getch(s[i])] == -1)
 83                 next[now][getch(s[i])] = newnode();
 84             now = next[now][getch(s[i])];
 85         }
 86         end[now]=true;
 87     }
 88     void build()
 89     {
 90         queue<int>Q;
 91         for(int i = 0;i < 4;i++)
 92             if(next[root][i] == -1)
 93                 next[root][i] = root;
 94             else
 95             {
 96                 fail[next[root][i]] = root;
 97                 Q.push(next[root][i]);
 98             }
 99         while(!Q.empty())
100         {
101             int now = Q.front();
102             Q.pop();
103             if(end[fail[now]]==true)
104                 end[now]=true;
105             for(int i = 0;i < 4;i++)
106             {
107                 if(next[now][i] == -1)
108                     next[now][i] = next[fail[now]][i];
109                 else
110                 {
111                     fail[next[now][i]] = next[fail[now]][i];
112                     Q.push(next[now][i]);
113                 }
114             }
115         }
116     }
117     Matrix getMatrix()
118     {
119         Matrix res = Matrix(L);
120         for(int i=0;i<L;i++)
121             for(int j=0;j<4;j++)
122                 if(end[next[i][j]]==false)
123                     res.mat[i][next[i][j]]++;
124         return res;
125     }
126 };
127 
128 Trie ac;
129 char buf[20];
130 
131 Matrix pow_M(Matrix a,int n)
132 {
133     Matrix ret = Matrix(a.n);
134     for(int i = 0; i < ret.n; i++)
135         ret.mat[i][i]=1;
136     Matrix tmp=a;
137     while(n)
138     {
139         if(n&1)ret=ret*tmp;
140         tmp=tmp*tmp;
141         n>>=1;
142     }
143     return ret;
144 }
145 
146 int main()
147 {
148     int n,m;
149     while(scanf("%d%d",&n,&m) != EOF)
150     {
151         ac.init();
152         for(int i=0;i<n;i++)
153         {
154             scanf("%s",buf);
155             ac.insert(buf);
156         }
157         ac.build();
158         Matrix a=ac.getMatrix();
159         a=pow_M(a,m);
160         int ans=0;
161         for(int i=0;i<a.n;i++)
162         {
163             ans=(ans+a.mat[0][i])%MOD;
164         }
165         printf("%d
",ans);
166     }
167     return 0;
168 }

View Code

6、HDU 2243 考研路茫茫——单词情结

这题和上题有些类似。但是需要求和。

所以给

矩阵增加一维，这样可以完美解决

题解here

  1 //============================================================================
  2 // Name        : HDU.cpp
  3 // Author      : 
  4 // Version     :
  5 // Copyright   : Your copyright notice
  6 // Description : Hello World in C++, Ansi-style
  7 //============================================================================
  8 
  9 #include <iostream>
 10 #include <stdio.h>
 11 #include <string.h>
 12 #include <algorithm>
 13 #include <queue>
 14 using namespace std;
 15 struct Matrix
 16 {
 17     unsigned long long mat[40][40];
 18     int n;
 19     Matrix(){}
 20     Matrix(int _n)
 21     {
 22         n=_n;
 23         for(int i=0;i<n;i++)
 24             for(int j=0;j<n;j++)
 25                 mat[i][j] = 0;
 26     }
 27     Matrix operator *(const Matrix &b)const
 28     {
 29         Matrix ret = Matrix(n);
 30         for(int i=0;i<n;i++)
 31             for(int j=0;j<n;j++)
 32                 for(int k=0;k<n;k++)
 33                     ret.mat[i][j]+=mat[i][k]*b.mat[k][j];
 34         return ret;
 35     }
 36 };
 37 unsigned long long pow_m(unsigned long long a,int n)
 38 {
 39     unsigned long long ret=1;
 40     unsigned long long tmp = a;
 41     while(n)
 42     {
 43         if(n&1)ret*=tmp;
 44         tmp*=tmp;
 45         n>>=1;
 46     }
 47     return ret;
 48 }
 49 Matrix pow_M(Matrix a,int n)
 50 {
 51     Matrix ret = Matrix(a.n);
 52     for(int i=0;i<a.n;i++)
 53         ret.mat[i][i] = 1;
 54     Matrix tmp = a;
 55     while(n)
 56     {
 57         if(n&1)ret=ret*tmp;
 58         tmp=tmp*tmp;
 59         n>>=1;
 60     }
 61     return ret;
 62 }
 63 struct Trie
 64 {
 65     int next[40][26],fail[40];
 66     bool end[40];
 67     int root,L;
 68     int newnode()
 69     {
 70         for(int i = 0;i < 26;i++)
 71             next[L][i] = -1;
 72         end[L++] = false;
 73         return L-1;
 74     }
 75     void init()
 76     {
 77         L = 0;
 78         root = newnode();
 79     }
 80     void insert(char buf[])
 81     {
 82         int len = strlen(buf);
 83         int now = root;
 84         for(int i = 0;i < len;i++)
 85         {
 86             if(next[now][buf[i]-'a'] == -1)
 87                 next[now][buf[i]-'a'] = newnode();
 88             now = next[now][buf[i]-'a'];
 89         }
 90         end[now] = true;
 91     }
 92     void build()
 93     {
 94         queue<int>Q;
 95         fail[root]=root;
 96         for(int i = 0;i < 26;i++)
 97             if(next[root][i] == -1)
 98                 next[root][i] = root;
 99             else
100             {
101                 fail[next[root][i]] = root;
102                 Q.push(next[root][i]);
103             }
104         while(!Q.empty())
105         {
106             int now = Q.front();
107             Q.pop();
108             if(end[fail[now]])end[now]=true;
109             for(int i = 0;i < 26;i++)
110                 if(next[now][i] == -1)
111                     next[now][i] = next[fail[now]][i];
112                 else
113                 {
114                     fail[next[now][i]] = next[fail[now]][i];
115                     Q.push(next[now][i]);
116                 }
117         }
118     }
119     Matrix getMatrix()
120     {
121         Matrix ret = Matrix(L+1);
122         for(int i = 0;i < L;i++)
123             for(int j = 0;j < 26;j++)
124                 if(end[next[i][j]]==false)
125                     ret.mat[i][next[i][j]] ++;
126         for(int i = 0;i < L+1;i++)
127             ret.mat[i][L] = 1;
128         return ret;
129     }
130     void debug()
131     {
132         for(int i = 0;i < L;i++)
133         {
134             printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
135             for(int j = 0;j < 26;j++)
136                 printf("%2d",next[i][j]);
137             printf("]
");
138         }
139     }
140 };
141 char buf[10];
142 Trie ac;
143 int main()
144 {
145 //    freopen("in.txt","r",stdin);
146 //    freopen("out.txt","w",stdout);
147     int n,L;
148     while(scanf("%d%d",&n,&L)==2)
149     {
150         ac.init();
151         for(int i = 0;i < n;i++)
152         {
153             scanf("%s",buf);
154             ac.insert(buf);
155         }
156         ac.build();
157         Matrix a = ac.getMatrix();
158         a = pow_M(a,L);
159         unsigned long long res = 0;
160         for(int i = 0;i < a.n;i++)
161             res += a.mat[0][i];
162         res--;
163 
164         /*
165          * f[n]=1 + 26^1 + 26^2 +...26^n
166          * f[n]=26*f[n-1]+1
167          * {f[n] 1} = {f[n-1] 1}[26 0;1 1]
168          * 数是f[L]-1;
169          * 此题的L<2^31.矩阵的幂不能是L+1次，否则就超时了
170          */
171         a = Matrix(2);
172         a.mat[0][0]=26;
173         a.mat[1][0] = a.mat[1][1] = 1;
174         a=pow_M(a,L);
175         unsigned long long ans=a.mat[1][0]+a.mat[0][0];
176         ans--;
177         ans-=res;
178         cout<<ans<<endl;
179     }
180     return 0;
181 }

View Code

7、POJ 1625 Censored!

AC自动机+DP+高精度

好题

题解here

//============================================================================
// Name        : POJ.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <queue>
#include <map>
using namespace std;
map<char,int>mp;
int N,M,P;
struct Matrix
{
    int mat[110][110];
    int n;
    Matrix(){}
    Matrix(int _n)
    {
        n=_n;
        for(int i = 0;i < n;i++)
            for(int j = 0;j < n;j++)
                mat[i][j] = 0;
    }
};
struct Trie
{
    int next[110][256],fail[110];
    bool end[110];
    int L,root;
    int newnode()
    {
        for(int i = 0;i < 256;i++)
            next[L][i] = -1;
        end[L++] = false;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][mp[buf[i]]] == -1)
                next[now][mp[buf[i]]] = newnode();
            now = next[now][mp[buf[i]]];
        }
        end[now] = true;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 256;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            if(end[fail[now]]==true)end[now]=true;
            for(int i = 0;i < 256;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    Matrix getMatrix()
    {
        Matrix res = Matrix(L);
        for(int i = 0;i < L;i++)
            for(int j = 0;j < N;j++)
                if(end[next[i][j]]==false)
                    res.mat[i][next[i][j]]++;
        return res;
    }
    void debug()
    {
        for(int i = 0;i < L;i++)
        {
            printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
            for(int j = 0;j < 26;j++)
                printf("%2d",next[i][j]);
            printf("]
");
        }
    }

};


/*
 * 高精度，支持乘法和加法
 */
struct BigInt
{
    const static int mod = 10000;
    const static int DLEN = 4;
    int a[600],len;
    BigInt()
    {
        memset(a,0,sizeof(a));
        len = 1;
    }
    BigInt(int v)
    {
        memset(a,0,sizeof(a));
        len = 0;
        do
        {
            a[len++] = v%mod;
            v /= mod;
        }while(v);
    }
    BigInt(const char s[])
    {
        memset(a,0,sizeof(a));
        int L = strlen(s);
        len = L/DLEN;
        if(L%DLEN)len++;
        int index = 0;
        for(int i = L-1;i >= 0;i -= DLEN)
        {
            int t = 0;
            int k = i - DLEN + 1;
            if(k < 0)k = 0;
            for(int j = k;j <= i;j++)
                t = t*10 + s[j] - '0';
            a[index++] = t;
        }
    }
    BigInt operator +(const BigInt &b)const
    {
        BigInt res;
        res.len = max(len,b.len);
        for(int i = 0;i <= res.len;i++)
            res.a[i] = 0;
        for(int i = 0;i < res.len;i++)
        {
            res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0);
            res.a[i+1] += res.a[i]/mod;
            res.a[i] %= mod;
        }
        if(res.a[res.len] > 0)res.len++;
        return res;
    }
    BigInt operator *(const BigInt &b)const
    {
        BigInt res;
        for(int i = 0; i < len;i++)
        {
            int up = 0;
            for(int j = 0;j < b.len;j++)
            {
                int temp = a[i]*b.a[j] + res.a[i+j] + up;
                res.a[i+j] = temp%mod;
                up = temp/mod;
            }
            if(up != 0)
                res.a[i + b.len] = up;
        }
        res.len = len + b.len;
        while(res.a[res.len - 1] == 0 &&res.len > 1)res.len--;
        return res;
    }
    void output()
    {
        printf("%d",a[len-1]);
        for(int i = len-2;i >=0 ;i--)
            printf("%04d",a[i]);
        printf("
");
    }
};
char buf[1010];
BigInt dp[2][110];
Trie ac;
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);

    while(scanf("%d%d%d",&N,&M,&P)==3)
    {
        gets(buf);
        gets(buf);
        mp.clear();
        int len = strlen(buf);
        for(int i = 0;i < len;i++)
            mp[buf[i]]=i;
        ac.init();
        for(int i = 0;i < P;i++)
        {
            gets(buf);
            ac.insert(buf);
        }
        ac.build();
//        ac.debug();
        Matrix a= ac.getMatrix();

//        for(int i = 0;i <a.n;i++)
//        {
//            for(int j=0;j<a.n;j++)printf("%d ",a.mat[i][j]);
//            cout<<endl;
//        }

        int now = 0;
        dp[now][0] = 1;
        for(int i = 1;i < a.n;i++)
            dp[now][i] = 0;
        for(int i = 0;i < M;i++)
        {
            now^=1;
            for(int j = 0;j < a.n;j++)
                dp[now][j] = 0;
            for(int j = 0;j < a.n;j++)
                for(int k = 0;k < a.n;k++)
                    if(a.mat[j][k] > 0)
                        dp[now][k] = dp[now][k]+dp[now^1][j]*a.mat[j][k];
//            for(int j = 0;j < a.n;j++)
//                dp[now][j].output();
        }
        BigInt ans = 0;
        for(int i = 0;i < a.n;i++)
            ans = ans + dp[now][i];
        ans.output();
    }
    return 0;
}

View Code

8、HDU 2825 Wireless Password

AC自动机+状态压缩DP

相当于在AC自动机上产生状态转移，然后进行dp

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;
const int MOD=20090717;
int n,m,k;
int dp[30][110][1<<10];
int num[5000];

struct Trie
{
    int next[110][26],fail[110],end[110];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 26;i++)
            next[L][i] = -1;
        end[L++] = 0;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[],int id)
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][buf[i]-'a']==-1)
                next[now][buf[i]-'a'] = newnode();
            now = next[now][buf[i]-'a'];
        }
        end[now] |= (1<<id);
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 26;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            end[now] |= end[fail[now]];
            for(int i = 0;i < 26;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    int solve()
    {
        //memset(dp,0,sizeof(dp));
        for(int i = 0;i <= n;i++)
            for(int j = 0;j < L;j++)
                for(int p = 0;p < (1<<m);p++)
                    dp[i][j][p]=0;
        dp[0][0][0] = 1;
        for(int i = 0;i < n;i++)
            for(int j = 0;j < L;j++)
                for(int p = 0;p< (1<<m);p++)
                    if(dp[i][j][p] > 0)
                    {
                        for(int x = 0;x < 26;x++)
                        {
                            int newi = i+1;
                            int newj = next[j][x];
                            int newp = (p|end[newj]);
                            dp[newi][newj][newp] += dp[i][j][p];
                            dp[newi][newj][newp]%=MOD;
                        }
                    }
        int ans = 0;
        for(int p = 0;p < (1<<m);p++)
        {
            if(num[p] < k)continue;
            for(int i = 0;i < L;i++)
            {
                ans = (ans + dp[n][i][p])%MOD;
            }
        }
        return ans;
    }
};
char buf[20];
Trie ac;
int main()
{
    for(int i=0;i<(1<<10);i++)
    {
        num[i] = 0;
        for(int j = 0;j < 10;j++)
            if(i & (1<<j))
                num[i]++;
    }
    while(scanf("%d%d%d",&n,&m,&k)==3)
    {
        if(n== 0 && m==0 &&k==0)break;
        ac.init();
        for(int i = 0;i < m;i++)
        {
            scanf("%s",buf);
            ac.insert(buf,i);
        }
        ac.build();
        printf("%d
",ac.solve());
    }
    return 0;
}

View Code

9、HDU 2296 Ring

需要输出字典序最小的解

在DP的时候加一个字符数组来记录就行了

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;

int a[110];
int dp[55][1110];
char str[55][1110][55];

bool cmp(char s1[],char s2[])
{
    int len1=strlen(s1);
    int len2=strlen(s2);
    if(len1 != len2)return len1 < len2;
    else return strcmp(s1,s2) < 0;
}

const int INF=0x3f3f3f3f;
struct Trie
{
    int next[1110][26],fail[1110],end[1110];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 26;i++)
            next[L][i] = -1;
        end[L++] = -1;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[],int id)
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][buf[i]-'a'] == -1)
                next[now][buf[i]-'a'] = newnode();
            now = next[now][buf[i]-'a'];
        }
        end[now] = id;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 26;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            for(int i = 0;i < 26;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    void solve(int n)
    {
        for(int i = 0;i <= n;i++)
            for(int j = 0;j < L;j++)
                dp[i][j] = -INF;
        dp[0][0] = 0;
        strcpy(str[0][0],"");
        char ans[55];
        strcpy(ans,"");
        int Max = 0;
        char tmp[55];
        for(int i = 0; i < n;i++)
            for(int j = 0;j < L;j++)
                if(dp[i][j]>=0)
                {
                    strcpy(tmp,str[i][j]);
                    int len = strlen(tmp);
                    for(int k = 0;k < 26;k++)
                    {
                        int nxt=next[j][k];
                        tmp[len] = 'a'+k;
                        tmp[len+1] = 0;
                        int tt = dp[i][j];
                        if(end[nxt] != -1)
                            tt+=a[end[nxt]];

                        if(dp[i+1][nxt]<tt || (dp[i+1][nxt]==tt && cmp(tmp,str[i+1][nxt])))
                        {
                            dp[i+1][nxt] = tt;
                            strcpy(str[i+1][nxt],tmp);
                            if(tt > Max ||(tt==Max && cmp(tmp,ans)))
                            {
                                Max = tt;
                                strcpy(ans,tmp);
                            }
                        }
                    }
                }
        printf("%s
",ans);
    }
};
char buf[60];
Trie ac;
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int T;
    int n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        ac.init();
        for(int i = 0;i < m;i++)
        {
            scanf("%s",buf);
            ac.insert(buf,i);
        }
        for(int i = 0;i < m;i++)
            scanf("%d",&a[i]);
        ac.build();
        ac.solve(n);
    }
    return 0;
}

View Code

10、HDU 2457 DNA repair

很简单的AC自动机+DP了

  1 //============================================================================
  2 // Name        : HDU.cpp
  3 // Author      : 
  4 // Version     :
  5 // Copyright   : Your copyright notice
  6 // Description : Hello World in C++, Ansi-style
  7 //============================================================================
  8 
  9 #include <iostream>
 10 #include <string.h>
 11 #include <stdio.h>
 12 #include <algorithm>
 13 #include <queue>
 14 using namespace std;
 15 const int INF = 0x3f3f3f3f;
 16 struct Trie
 17 {
 18     int next[1010][4],fail[1010];
 19     bool end[1010];
 20     int root,L;
 21     int newnode()
 22     {
 23         for(int i = 0;i < 4;i++)
 24             next[L][i] = -1;
 25         end[L++] = false;
 26         return L-1;
 27     }
 28     void init()
 29     {
 30         L = 0;
 31         root = newnode();
 32     }
 33     int getch(char ch)
 34     {
 35         if(ch == 'A')return 0;
 36         else if(ch == 'C')return 1;
 37         else if(ch == 'G')return 2;
 38         else if(ch == 'T')return 3;
 39     }
 40     void insert(char buf[])
 41     {
 42         int len = strlen(buf);
 43         int now = root;
 44         for(int i = 0;i < len;i++)
 45         {
 46             if(next[now][getch(buf[i])] == -1)
 47                 next[now][getch(buf[i])] = newnode();
 48             now = next[now][getch(buf[i])];
 49         }
 50         end[now] = true;
 51     }
 52     void build()
 53     {
 54         queue<int>Q;
 55         fail[root] = root;
 56         for(int i = 0;i < 4;i++)
 57             if(next[root][i] == -1)
 58                 next[root][i] = root;
 59             else
 60             {
 61                 fail[next[root][i]] = root;
 62                 Q.push(next[root][i]);
 63             }
 64         while(!Q.empty())
 65         {
 66             int now = Q.front();
 67             Q.pop();
 68             if(end[fail[now]])end[now] = true;//这里不要忘记
 69             for(int i = 0;i < 4;i++)
 70                 if(next[now][i] == -1)
 71                     next[now][i] = next[fail[now]][i];
 72                 else
 73                 {
 74                     fail[next[now][i]] = next[fail[now]][i];
 75                     Q.push(next[now][i]);
 76                 }
 77         }
 78     }
 79     int dp[1010][1010];
 80     int solve(char buf[])
 81     {
 82         int len = strlen(buf);
 83         for(int i = 0;i <= len;i++)
 84             for(int j = 0;j < L;j++)
 85                 dp[i][j] = INF;
 86         dp[0][root] = 0;
 87         for(int i = 0;i < len;i++)
 88             for(int j = 0;j < L;j++)
 89                 if(dp[i][j] < INF)
 90                 {
 91                     for(int k = 0;k < 4;k++)
 92                     {
 93                         int news = next[j][k];
 94                         if(end[news])continue;
 95                         int tmp;
 96                         if( k == getch(buf[i]))tmp = dp[i][j];
 97                         else tmp = dp[i][j] + 1;
 98                         dp[i+1][news] = min(dp[i+1][news],tmp);
 99                     }
100                 }
101         int ans = INF;
102         for(int j = 0;j < L;j++)
103             ans = min(ans,dp[len][j]);
104         if(ans == INF)ans = -1;
105         return ans;
106     }
107 
108 };
109 char buf[1010];
110 Trie ac;
111 int main()
112 {
113     int n;
114     int iCase = 0;
115     while ( scanf("%d",&n) == 1 && n)
116     {
117         iCase++;
118         ac.init();
119         while(n--)
120         {
121             scanf("%s",buf);
122             ac.insert(buf);
123         }
124         ac.build();
125         scanf("%s",buf);
126         printf("Case %d: %d
",iCase,ac.solve(buf));
127     }
128     return 0;
129 }

View Code

11、ZOJ 3228 Searching the String

这题需要查询两种，一种是可重叠，一种是不可重叠的。

找模式串在目标串中的出现次数。

加一个数组记录上一次出现的位置，然后就可以求出不可重叠的了

  1 //============================================================================
  2 // Name        : ZOJ.cpp
  3 // Author      : 
  4 // Version     :
  5 // Copyright   : Your copyright notice
  6 // Description : Hello World in C++, Ansi-style
  7 //============================================================================
  8 
  9 #include <iostream>
 10 #include <stdio.h>
 11 #include <string.h>
 12 #include <algorithm>
 13 #include <queue>
 14 using namespace std;
 15 
 16 struct Trie
 17 {
 18     int next[600010][26],fail[600010],deep[600010];
 19     int root,L;
 20     int newnode()
 21     {
 22         for(int i = 0;i < 26;i++)
 23             next[L][i] = -1;
 24         L++;
 25         return L-1;
 26     }
 27     void init()
 28     {
 29         L = 0;
 30         root = newnode();
 31         deep[root] = 0;
 32     }
 33     int insert(char buf[])
 34     {
 35         int len = strlen(buf);
 36         int now = root;
 37         for(int i = 0;i < len;i++)
 38         {
 39             if(next[now][buf[i]-'a'] == -1)
 40             {
 41                 next[now][buf[i]-'a'] = newnode();
 42                 deep[ next[now][buf[i]-'a'] ] = i+1;
 43             }
 44             now = next[now][buf[i]-'a'];
 45         }
 46         return now;
 47     }
 48     void build()
 49     {
 50         queue<int>Q;
 51         fail[root] = root;
 52         for(int i = 0;i < 26;i++)
 53             if(next[root][i] == -1)
 54                 next[root][i] = root;
 55             else
 56             {
 57                 fail[next[root][i]] = root;
 58                 Q.push(next[root][i]);
 59             }
 60         while(!Q.empty())
 61         {
 62             int now = Q.front();
 63             Q.pop();
 64             for(int i = 0;i < 26;i++)
 65                 if(next[now][i] == -1)
 66                     next[now][i] = next[fail[now]][i];
 67                 else
 68                 {
 69                     fail[next[now][i]] = next[fail[now]][i];
 70                     Q.push(next[now][i]);
 71                 }
 72         }
 73     }
 74     int cnt[600010][2];
 75     int last[600010];
 76     void query(char buf[])
 77     {
 78         int len = strlen(buf);
 79         int now = root;
 80         memset(cnt,0,sizeof(cnt));
 81         memset(last,-1,sizeof(last));
 82         for(int i = 0;i < len;i++)
 83         {
 84             now = next[now][buf[i]-'a'];
 85             int temp = now;
 86             while(temp != root)
 87             {
 88                 cnt[temp][0]++;
 89                 if(i - last[temp] >= deep[temp])
 90                 {
 91                     last[temp] = i;
 92                     cnt[temp][1]++;
 93                 }
 94                 temp = fail[temp];
 95             }
 96         }
 97     }
 98 };
 99 Trie ac;
100 char str[100010];
101 char buf[20];
102 int typ[100010],pos[100010];
103 int main()
104 {
105 //    freopen("in.txt","r",stdin);
106 //    freopen("out.txt","w",stdout);
107     int n;
108     int iCase = 0;
109     while(scanf("%s",str)==1)
110     {
111         iCase++;
112         printf("Case %d
",iCase);
113         scanf("%d",&n);
114         ac.init();
115         for(int i = 0;i < n;i++)
116         {
117             scanf("%d%s",&typ[i],buf);
118             pos[i]=ac.insert(buf);
119         }
120         ac.build();
121         ac.query(str);
122         for(int i = 0;i < n;i++)
123             printf("%d
",ac.cnt[pos[i]][typ[i]]);
124         printf("
");
125     }
126     return 0;
127 }

View Code

12、HDU 3341 Lost's revenge

这题主要是状态的表示，就是记录ACGT出现的次数。

然后这个ACGT次数表示的时候，状态要转化。

题解here

  1 //============================================================================
  2 // Name        : HDU.cpp
  3 // Author      : 
  4 // Version     :
  5 // Copyright   : Your copyright notice
  6 // Description : Hello World in C++, Ansi-style
  7 //============================================================================
  8 
  9 #include <iostream>
 10 #include <string.h>
 11 #include <stdio.h>
 12 #include <algorithm>
 13 #include <queue>
 14 using namespace std;
 15 const int INF = 0x3f3f3f3f;
 16 struct Trie
 17 {
 18     int next[510][4],fail[510];
 19     int end[510];
 20     int root,L;
 21     int newnode()
 22     {
 23         for(int i = 0;i < 4;i++)
 24             next[L][i] = -1;
 25         end[L++] = 0;
 26         return L-1;
 27     }
 28     void init()
 29     {
 30         L = 0;
 31         root = newnode();
 32     }
 33     int getch(char ch)
 34     {
 35         if(ch == 'A')return 0;
 36         else if(ch == 'C')return 1;
 37         else if(ch == 'G')return 2;
 38         else return 3;
 39     }
 40     void insert(char buf[])
 41     {
 42         int len = strlen(buf);
 43         int now = root;
 44         for(int i = 0;i < len;i++)
 45         {
 46             if(next[now][getch(buf[i])] == -1)
 47                 next[now][getch(buf[i])] = newnode();
 48             now = next[now][getch(buf[i])];
 49         }
 50         end[now] ++;
 51     }
 52     void build()
 53     {
 54         queue<int>Q;
 55         fail[root] = root;
 56         for(int i = 0;i < 4;i++)
 57             if(next[root][i] == -1)
 58                 next[root][i] = root;
 59             else
 60             {
 61                 fail[next[root][i]] = root;
 62                 Q.push(next[root][i]);
 63             }
 64         while(!Q.empty())
 65         {
 66             int now = Q.front();
 67             Q.pop();
 68             end[now] += end[fail[now]];/********/
 69             for(int i = 0;i < 4;i++)
 70                 if(next[now][i] == -1)
 71                     next[now][i] = next[fail[now]][i];
 72                 else
 73                 {
 74                     fail[next[now][i]] = next[fail[now]][i];
 75                     Q.push(next[now][i]);
 76                 }
 77         }
 78     }
 79     int dp[510][11*11*11*11+5];
 80     int bit[4];
 81     int num[4];
 82     int solve(char buf[])
 83     {
 84         int len = strlen(buf);
 85         memset(num,0,sizeof(num));
 86         for(int i = 0;i < len;i++)
 87             num[getch(buf[i])]++;
 88         bit[0] = (num[1]+1)*(num[2]+1)*(num[3]+1);
 89         bit[1] = (num[2]+1)*(num[3]+1);
 90         bit[2] = (num[3]+1);
 91         bit[3] = 1;
 92         memset(dp,-1,sizeof(dp));
 93         dp[root][0] = 0;
 94         for(int A = 0;A <= num[0];A++)
 95             for(int B = 0;B <= num[1];B++)
 96                 for(int C = 0;C <= num[2];C++)
 97                     for(int D = 0;D <= num[3];D++)
 98                     {
 99                         int s = A*bit[0] + B*bit[1] + C*bit[2] + D*bit[3];
100                         for(int i = 0;i < L;i++)
101                             if(dp[i][s] >= 0)
102                             {
103                                 for(int k = 0;k < 4;k++)
104                                 {
105                                     if(k == 0 && A == num[0])continue;
106                                     if(k == 1 && B == num[1])continue;
107                                     if(k == 2 && C == num[2])continue;
108                                     if(k == 3 && D == num[3])continue;
109                                     dp[next[i][k]][s+bit[k]] = max(dp[next[i][k]][s+bit[k]],dp[i][s]+end[next[i][k]]);
110                                 }
111                             }
112                     }
113         int ans = 0;
114         int status = num[0]*bit[0] + num[1]*bit[1] + num[2]*bit[2] + num[3]*bit[3];
115         for(int i = 0;i < L;i++)
116             ans = max(ans,dp[i][status]);
117         return ans;
118     }
119 };
120 char buf[50];
121 Trie ac;
122 int main()
123 {
124 //    freopen("in.txt","r",stdin);
125 //    freopen("out.txt","w",stdout);
126     int n;
127     int iCase = 0;
128     while( scanf("%d",&n) == 1 && n)
129     {
130         iCase++;
131         ac.init();
132         for(int i = 0;i < n;i++)
133         {
134             scanf("%s",buf);
135             ac.insert(buf);
136         }
137         ac.build();
138         scanf("%s",buf);
139         printf("Case %d: %d
",iCase,ac.solve(buf));
140     }
141     return 0;
142 }

View Code

13、HDU 3247 Resource Archiver

使用最短路预处理出状态的转移。这样可以优化很多

  1 //============================================================================
  2 // Name        : HDU.cpp
  3 // Author      : 
  4 // Version     :
  5 // Copyright   : Your copyright notice
  6 // Description : Hello World in C++, Ansi-style
  7 //============================================================================
  8 
  9 #include <iostream>
 10 #include <stdio.h>
 11 #include <string.h>
 12 #include <algorithm>
 13 #include <queue>
 14 using namespace std;
 15 
 16 const int INF = 0x3f3f3f3f;
 17 
 18 struct Trie
 19 {
 20     int next[60010][2],fail[60010],end[60010];
 21     int root,L;
 22     int newnode()
 23     {
 24         for(int i = 0;i < 2;i++)
 25             next[L][i] = -1;
 26         end[L++] = 0;
 27         return L-1;
 28     }
 29     void init()
 30     {
 31         L = 0;
 32         root = newnode();
 33     }
 34     void insert(char buf[],int id)
 35     {
 36         int len = strlen(buf);
 37         int now = root;
 38         for(int i = 0;i < len ;i++)
 39         {
 40             if(next[now][buf[i]-'0'] == -1)
 41                 next[now][buf[i]-'0'] = newnode();
 42             now = next[now][buf[i]-'0'];
 43         }
 44         end[now] = id;
 45     }
 46     void build()
 47     {
 48         queue<int>Q;
 49         fail[root] = root;
 50         for(int i = 0;i < 2;i++)
 51             if(next[root][i] == -1)
 52                 next[root][i] = root;
 53             else
 54             {
 55                 fail[next[root][i]] = root;
 56                 Q.push(next[root][i]);
 57             }
 58         while( !Q.empty() )
 59         {
 60             int now = Q.front();
 61             Q.pop();
 62             if(end[fail[now]] == -1)end[now] = -1;
 63             else end[now] |= end[fail[now]];
 64             for(int i = 0;i < 2;i++)
 65                 if(next[now][i] == -1)
 66                     next[now][i] = next[fail[now]][i];
 67                 else
 68                 {
 69                     fail[next[now][i]] = next[fail[now]][i];
 70                     Q.push(next[now][i]);
 71                 }
 72         }
 73     }
 74     int g[11][11];
 75     int dp[1025][11];
 76     int cnt;
 77     int pos[11];
 78     int dis[60010];
 79 
 80 
 81     void bfs(int k)
 82     {
 83         queue<int>q;
 84         memset(dis,-1,sizeof(dis));
 85         dis[pos[k]] = 0;
 86         q.push(pos[k]);
 87         while(!q.empty())
 88         {
 89             int now = q.front();
 90             q.pop();
 91             for(int i = 0; i< 2;i++)
 92             {
 93                 int tmp = next[now][i];
 94                 if(dis[tmp]<0 && end[tmp] >= 0)
 95                 {
 96                     dis[tmp] = dis[now] + 1;
 97                     q.push(tmp);
 98                 }
 99             }
100         }
101         for(int i = 0;i < cnt;i++)
102             g[k][i] = dis[pos[i]];
103     }
104 
105 
106     int solve(int n)
107     {
108 
109         pos[0] = 0;
110         cnt = 1;
111         for(int i = 0;i < L;i++)
112             if(end[i] > 0)
113                 pos[cnt++] = i;
114         for(int i = 0; i < cnt;i++)
115             bfs(i);
116 
117         for(int i = 0;i < (1<<n);i++)
118             for(int j = 0;j < cnt;j++)
119                 dp[i][j] = INF;
120         dp[0][0] = 0;
121         for(int i = 0;i <(1<<n);i++)
122             for(int j = 0;j < cnt;j++)
123                 if(dp[i][j]<INF)
124                 {
125                     for(int k = 0;k < cnt;k++)
126                     {
127                         if(g[j][k] < 0)continue;
128                         if( j == k)continue;
129                         dp[i|end[pos[k]]][k] = min(dp[i|end[pos[k]]][k],dp[i][j]+g[j][k]);
130                     }
131                 }
132         int ans = INF;
133         for(int j = 0;j < cnt;j++)
134             ans = min(ans,dp[(1<<n)-1][j]);
135         return ans;
136     }
137 };
138 Trie ac;
139 char buf[1010];
140 
141 int main()
142 {
143 //    freopen("in.txt","r",stdin);
144 //    freopen("out.txt","w",stdout);
145     int n,m;
146     while(scanf("%d%d",&n,&m) == 2)
147     {
148         if(n == 0 && m == 0)break;
149         ac.init();
150         for(int i = 0;i < n;i++)
151         {
152             scanf("%s",buf);
153             ac.insert(buf,1<<i);
154         }
155         for(int i = 0;i < m;i++)
156         {
157             scanf("%s",buf);
158             ac.insert(buf,-1);
159         }
160         ac.build();
161         printf("%d
",ac.solve(n));
162     }
163     return 0;
164 }

View Code

14、ZOJ 3494 BCD Code

这道题很神，数位DP和AC自动机结合，太强大了。

题解here

  1 //============================================================================
  2 // Name        : ZOJ.cpp
  3 // Author      : 
  4 // Version     :
  5 // Copyright   : Your copyright notice
  6 // Description : Hello World in C++, Ansi-style
  7 //============================================================================
  8 
  9 #include <iostream>
 10 #include <stdio.h>
 11 #include <string.h>
 12 #include <algorithm>
 13 #include <queue>
 14 using namespace std;
 15 
 16 struct Trie
 17 {
 18     int next[2010][2],fail[2010];
 19     bool end[2010];
 20     int root,L;
 21     int newnode()
 22     {
 23         for(int i = 0;i < 2;i++)
 24             next[L][i] = -1;
 25         end[L++] = false;
 26         return L-1;
 27     }
 28     void init()
 29     {
 30         L = 0;
 31         root = newnode();
 32     }
 33     void insert(char buf[])
 34     {
 35         int len = strlen(buf);
 36         int now = root;
 37         for(int i = 0;i < len ;i++)
 38         {
 39             if(next[now][buf[i]-'0'] == -1)
 40                 next[now][buf[i]-'0'] = newnode();
 41             now = next[now][buf[i]-'0'];
 42         }
 43         end[now] = true;
 44     }
 45     void build()
 46     {
 47         queue<int>Q;
 48         fail[root] = root;
 49         for(int i = 0;i < 2;i++)
 50             if(next[root][i] == -1)
 51                 next[root][i] = root;
 52             else
 53             {
 54                 fail[next[root][i]] = root;
 55                 Q.push(next[root][i]);
 56             }
 57         while(!Q.empty())
 58         {
 59             int now = Q.front();
 60             Q.pop();
 61             if(end[fail[now]])end[now] = true;
 62             for(int i = 0;i < 2;i++)
 63                 if(next[now][i] == -1)
 64                     next[now][i] = next[fail[now]][i];
 65                 else
 66                 {
 67                     fail[next[now][i]] = next[fail[now]][i];
 68                     Q.push(next[now][i]);
 69                 }
 70         }
 71     }
 72 };
 73 Trie ac;
 74 
 75 int bcd[2010][10];
 76 int change(int pre,int num)
 77 {
 78     if(ac.end[pre])return -1;
 79     int cur = pre;
 80     for(int i = 3;i >= 0;i--)
 81     {
 82         if(ac.end[ac.next[cur][(num>>i)&1]])return -1;
 83         cur = ac.next[cur][(num>>i)&1];
 84     }
 85     return cur;
 86 }
 87 void pre_init()
 88 {
 89     for(int i = 0;i <ac.L;i++)
 90         for(int j = 0;j <10;j++)
 91             bcd[i][j] = change(i,j);
 92 }
 93 const int MOD = 1000000009;
 94 long long dp[210][2010];
 95 int bit[210];
 96 
 97 long long dfs(int pos,int s,bool flag,bool z)
 98 {
 99     if(pos == -1)return 1;
100     if(!flag && dp[pos][s]!=-1)return dp[pos][s];
101     long long ans = 0;
102     if(z)
103     {
104         ans += dfs(pos-1,s,flag && bit[pos]==0,true);
105         ans %= MOD;
106     }
107     else
108     {
109         if(bcd[s][0]!=-1)ans += dfs(pos-1,bcd[s][0],flag && bit[pos]==0,false);
110         ans %= MOD;
111     }
112     int end = flag?bit[pos]:9;
113     for(int i = 1;i<=end;i++)
114     {
115         if(bcd[s][i]!=-1)
116         {
117             ans += dfs(pos-1,bcd[s][i],flag&&i==end,false);
118             ans %=MOD;
119         }
120     }
121     if(!flag && !z)dp[pos][s] = ans;
122     return ans;
123 }
124 
125 long long calc(char s[])
126 {
127     int len = strlen(s);
128     for(int i = 0;i < len;i++)
129         bit[i] = s[len-1-i]-'0';
130     return dfs(len-1,0,1,1);
131 }
132 char str[210];
133 int main()
134 {
135 //    freopen("in.txt","r",stdin);
136 //    freopen("out.txt","w",stdout);
137     int T;
138     scanf("%d",&T);
139     int n;
140     while(T--)
141     {
142         ac.init();
143         scanf("%d",&n);
144         for(int i = 0;i < n;i++)
145         {
146             scanf("%s",str);
147             ac.insert(str);
148         }
149         ac.build();
150         pre_init();
151         memset(dp,-1,sizeof(dp));
152         int ans = 0;
153         scanf("%s",str);
154         int len = strlen(str);
155         for(int i = len -1;i >=0;i--)
156         {
157             if(str[i]>'0')
158             {
159                 str[i]--;
160                 break;
161             }
162             else str[i] = '9';
163         }
164         ans -= calc(str);
165         ans %=MOD;
166         scanf("%s",str);
167         ans += calc(str);
168         ans %=MOD;
169         if(ans < 0)ans += MOD;
170         printf("%d
",ans);
171     }
172     return 0;
173 }

View Code

相关阅读:
HBase 解决HBase Client无法连接远程HBase Server问题
 HBase的介绍与安装教程（CentOS下单机版安装、hbase shell基本用法）
SpringBoot 使用Phoenix操作HBase教程2（使用JdbcTemplate）
HBase Phoenix的安装使用教程1（基本介绍、安装部署）
InfluxDB时序数据库的安装使用教程2（安装部署、命令行基本用法）
SpringBoot 使用hbaseclient操作HBase教程1（基本用法）
SpringBoot 使用hbaseclient操作HBase教程2（过滤器Filter）
HBase Phoenix的安装使用教程3（SCHEMA的启用、操作、关闭）
C#遍历enum类型
 arcengine 矢量要素编辑(转载)
原文地址：https://www.cnblogs.com/13224ACMer/p/4873257.html