hdu 2243 考研路茫茫——单词情结 ac自动机+矩阵快速幂

链接:http://acm.hdu.edu.cn/showproblem.php?pid=2243

题意:给定N(1<= N < 6)个长度不超过5的词根，问长度不超过L(L <2³¹)的单词中至少含有一个词根的单词个数；结果mod 2⁶⁴.

基础:poj 2778DNA 序列求的是给定长度不含模式串的合法串的个数；串长度相当，都到了int上界了；

1.mod 2⁶⁴直接使用unsigned long long自然溢出即可；说的有些含蓄。。并且也容易想到是直接使用内置类型，要不然高精度的mod 2^64要跪；

2.本题是poj 2778的升级版，里面求的是范围内，并不是特定的长度，并且还是含有至少一个；那么直接从反面求解，即用总个数-不含有词根的个数；

3.对矩阵的操作，实现对连续离散指数求和; 对矩阵增加一列，全部为1，增加的行除了右下角为1外，其他均为0；

$egin{pmatrix}
ARightarrow egin{pmatrix}
A  & 1\
0 & 1
end{pmatrix}
end{pmatrix}$

$ans = sum limits_{i=0}^{ m size}matleft [ 0 ight ]left [ i ight ]$

同理求解:$sum limits_{i=0}^{ m size}26^i$只需建立二维矩阵

egin{vmatrix}
26 & 1\
0 & 1
end{vmatrix}之后直接矩阵快速幂即可；

// 15ms 1632k
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
#include<stack>
#include<set>
#include<map>
#include<queue>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define inf 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1|1
#define sqr(a) (a)*(a)
typedef pair<int,int> PII;
#define A first
#define B second
#define MK make_pair
typedef __int64 ll;
typedef unsigned long long ull;
template<typename T>
void read1(T &m)
{
    T x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>9) out(a/10);
    putchar(a%10+'0');
}
int T,kase = 0,i,j,k,n,m,top;
#define mod 100000
struct Matrix{
    ull d[37][37];
    int sz;//不能也弄成ull,否则int => ull会出错
    Matrix(int r){MS0(d); sz = r;}
    void init(){
        for(int i = 0;i < sz;i++)
            d[i][i] = 1;
    }
    Matrix operator*(const Matrix& a)const{
        Matrix ans(a.sz);
        for(int i = 0;i < sz;i++)
        for(int j = 0;j < sz;j++){
            for(int k = 0;k < sz;k++)
                ans.d[i][j] += d[i][k]*a.d[k][j];
        }
        return ans;
    }
    void debug(){
        rep0(i,0,sz){
            rep0(j,0,sz) cout<<d[i][j]<<" ";
            cout<<endl;
        }
    }
};
Matrix Pow(Matrix a,ll m)
{
    Matrix ans(a.sz);
    ans.init();
    while(m){
        if(m&1) ans = ans*a;
        a = a*a;
        m >>= 1;
    }
    return ans;
}
const int sigma_size = 26;
const int maxn = 37;
struct Aho_Corasick{
    int ch[maxn][sigma_size];
    int val[maxn],f[maxn],last[maxn],cnt[maxn];
    int sz;
    map<string,int> ms;
    Aho_Corasick(){}
    void init(){sz = 1;val[0] = 0; MS0(ch[0]);MS0(cnt);ms.clear();}
    int idx(char c){return c-'a';}
    void Insert(char *s,int v){
        int u = 0,n = strlen(s);
        for(int i = 0;i < n;i++){
            int c = idx(s[i]);
            if(!ch[u][c]){
                MS0(ch[sz]);
                val[sz] = 0;
                ch[u][c] = sz++;
            }
            u = ch[u][c];
        }
        val[u] = v;
    }
    void getFail(){
        queue<int> q;
        f[0] = 0;
        for(int c = 0;c < sigma_size;c++){ //初始化队列
            int u = ch[0][c];
            if(u) { f[u] = 0; q.push(u); last[u] = 0;}
        }
        while(!q.empty()){
            int r = q.front();q.pop();
            for(int c = 0;c < sigma_size;c++){
                int u = ch[r][c];
                if(!u) {ch[r][c] = ch[f[r]][c]; continue;}//实现压缩
                q.push(u);
                int v = f[r];
                while(v && !ch[v][c]) v = f[v];
                f[u] = ch[v][c];
                last[u] = val[f[u]]?f[u]:last[f[u]];
            }
        }
    }
    Matrix Find(){
        Matrix ans(sz+1);
        for(int i = 0;i < sz;i++){
            if(val[i] || last[i]) continue;
            for(int j = 0;j < sigma_size;j++){
                int v = ch[i][j];
                if(val[v] || last[v]) continue;
                ans.d[i][v]++;
            }
            ans.d[i][sz] = 1;//添加一列保存前面的sigma和;
        }
        ans.d[sz][sz] = 1;//右下角
        return ans;
    }
}ac;
char p[15];
int main()
{
    ull n,L;
    //freopen("data.txt","r",stdin);
    while(scanf("%I64d%I64d",&n,&L) == 2) {
        ac.init();
        rep1(i,1,n){
            scanf("%s",p);
            ac.Insert(p,i);
        }
        ac.getFail();
        Matrix ans = ac.Find();
        ans = Pow(ans,L);
        ull cnt = 0;
        rep0(i,0,ans.sz) cnt += ans.d[0][i];
        Matrix a(2);//求总和
        a.d[0][0] = 26,a.d[0][1] = a.d[1][1] = 1;
        a = Pow(a,L);
        printf("%I64u
",a.d[0][0]+a.d[0][1]-cnt);
    }
    return 0;
}