HDU 2825 Wireless Password (AC自己主动机，DP)

http://acm.hdu.edu.cn/showproblem.php?

pid=2825

Wireless Password
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4560    Accepted Submission(s): 1381

Problem Description

Liyuan lives in a old apartment. One day, he suddenly found that there was a wireless network in the building. Liyuan did not know the password of the network, but he got some important information from his neighbor. He knew the password consists only of lowercase letters 'a'-'z', and he knew the length of the password. Furthermore, he got a magic word set, and his neighbor told him that the password included at least k words of the magic word set (the k words in the password possibly overlapping).

For instance, say that you know that the password is 3 characters long, and the magic word set includes 'she' and 'he'. Then the possible password is only 'she'.

Liyuan wants to know whether the information is enough to reduce the number of possible passwords. To answer this, please help him write a program that determines the number of possible passwords.

 

Input

There will be several data sets. Each data set will begin with a line with three integers n m k. n is the length of the password (1<=n<=25), m is the number of the words in the magic word set(0<=m<=10), and the number k denotes that the password included at least k words of the magic set. This is followed by m lines, each containing a word of the magic set, each word consists of between 1 and 10 lowercase letters 'a'-'z'. End of input will be marked by a line with n=0 m=0 k=0, which should not be processed.

 

Output

For each test case, please output the number of possible passwords MOD 20090717.

 

Sample Input

10 2 2
hello 
world 
4 1 1
icpc 
10 0 0
0 0 0

 

Sample Output

2
1
14195065

 

Source

2009 Multi-University Training Contest 1 - Host by TJU

 

题意：

给出m个模式串。要求构造一长度为n的文本串，至少包含k种模式串，求有多少种可能的模式串。

分析：

m个模式串构建AC自己主动机。然后要在这AC自己主动机中走n步。至少经过k个单词结点。由于m<=10,显然能够用状压表示已经有哪几个单词结点。用dp[i][j][k]表示走了i步到AC自己主动机中的第j个结点，单词状态为k。由计数原理可推出状态转移方程：dp[i][j][k]=sum(dp[i-1][last_j][last_k]),last_j表示能够抵达第j个结点的上一个结点。last_k表示上一步的状态;由于第i步仅仅和第i-1步有关。所以能够用滚动数组优化空间。

/*
 *
 * Author : fcbruce <fcbruce8964@gmail.com>
 *
 * Time : Thu 20 Nov 2014 10:01:45 AM CST
 *
 */
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10

#ifdef _WIN32
  #define lld "%I64d"
#else
  #define lld "%lld"
#endif

#define maxm 
#define maxn 255

using namespace std;

const int mod = 20090717;

int dp[2][maxn][1024];

inline int add(int a,int b)
{
  return (a+b)%mod;
}

int q[maxn];

const int maxsize = 26;
struct ACauto
{
  int ch[maxn][maxsize];
  int val[maxn],nex[maxn],last[maxn];
  int sz;

  ACauto()
  {
    memset(ch[0],0,sizeof ch[0]);
    val[0]=0;
    sz=1;
  }

  void clear()
  {
    memset(ch[0],0,sizeof ch[0]);
    val[0]=0;
    sz=1;
  }

  int idx(char c)
  {
    return c-'a';
  }

  void insert(const char *s,int v)
  {
    int u=0;
    for (int i=0;s[i]!='';i++)
    {
      int c=idx(s[i]);
      if (ch[u][c]==0)
      {
        memset(ch[sz],0,sizeof ch[sz]);
        val[sz]=0;
        ch[u][c]=sz++;
      }
      u=ch[u][c];
    }
    val[u]=v;
  }

  void get_fail()
  {
    int f=0,r=-1;
    nex[0]=0;
    for (int c=0;c<maxsize;c++)
    {
      int u=ch[0][c];
      if (u!=0)
      {
        nex[u]=0;
        q[++r]=u;
        last[u]=0;
      }
    }

    while (f<=r)
    {
      int x=q[f++];
      for (int c=0;c<maxsize;c++)
      {
        int u=ch[x][c];
        if (u==0)
        {
          ch[x][c]=ch[nex[x]][c];
          continue;
        }
        q[++r]=u;
        int v=nex[x];
        nex[u]=ch[v][c];
        val[u]|=val[nex[u]];
      }
    }
  }

  int calc(int x)
  {
    int cnt=0;
    for (int i=0;i<32;i++)
      if (x&(1<<i)) cnt++;
    return cnt;
  }
  
  int DP(int l,int m,int k)
  {
    memset(dp,0,sizeof dp);
    dp[0][0][0]=1;
    int x=1;
    for (int i=0;i<l;i++,x^=1)
    {
      memset(dp[x],0,sizeof dp[x]);
      for (int j=0;j<sz;j++)
      {
        for (int s=0;s<m;s++)
        {
          if (dp[x^1][j][s]==0) continue;
          for (int c=0;c<maxsize;c++)
          {
            int &cur=dp[x][ch[j][c]][s|val[ch[j][c]]];
            cur=add(cur,dp[x^1][j][s]);
          }
        }
      }
    }

    int total=0;

    for (int i=0;i<m;i++)
    {
      if (calc(i)<k) continue;
      for (int j=0;j<sz;j++)
        total=add(total,dp[x^1][j][i]);
    }

    return total;
  }
}acauto;

char str[16];

int main()
{
#ifdef FCBRUCE
  freopen("/home/fcbruce/code/t","r",stdin);
#endif // FCBRUCE

  int n,m,k;

  while (scanf("%d%d%d",&n,&m,&k),n||m||k)
  {
    acauto.clear();
    for (int i=0;i<m;i++)
    {
      scanf("%s",str);
      acauto.insert(str,1<<i);
    }

    acauto.get_fail();

    printf("%d
",acauto.DP(n,1<<m,k));
  }


  return 0;
}