HDU2243-AC自动机+矩阵

题意：中文题

题解思路：首先要做这题得先去做POJ2778，然后你知道那么长度不超过m的方案数就是26+26^2+…+26^m减去矩阵(M+M^2+…M^m)第一行的总和,因为该题取摸2^64直接用

unsigned long long 即可

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
typedef unsigned long long ul;
const int mx = 105;
struct mat{
    ul a[35][35];
};
mat M,I;
struct trie{
    int ch[mx][26];
    int f[mx];
    int v[mx];
    int sz;
    void init(){
        memset(ch[0],0,sizeof(ch[0]));
        memset(f,0,sizeof(f));
        memset(v,0,sizeof(v));
        sz = 1;
    }
    int idx(char c){
        return c-'a';
    }
    void insert(char *s){
        int len = strlen(s);
        int u = 0;
        for(int i = 0; i < len; i++){
            int d = idx(s[i]);
            if(!ch[u][d]){
                memset(ch[sz],0,sizeof(ch[sz]));
                ch[u][d] = sz++;
            }
            u = ch[u][d];
        }
        v[u] = 1;
    }
    void getfail(){
        queue<int>q;
        for(int d = 0; d < 26; d++){
            if(!ch[0][d])
                continue;
            int ret = ch[0][d];
            f[ret] = 0;
            q.push(ret);
        }
        while(!q.empty()){
            int u = q.front();
            //cout<<u<<endl;
            q.pop();
            if(v[f[u]])
                 v[u] = 1;
            for(int d = 0; d < 26; d++){

                if(!ch[u][d]){
                    ch[u][d] = ch[f[u]][d];
                    continue;
                }
                int ret = ch[u][d];
                f[ret] = ch[f[u]][d];
                q.push(ret);
            }
        }
    }
    void getmat(){
        memset(I.a,0,sizeof(I.a));
        memset(M.a,0,sizeof(M.a));
        for(int u = 0; u < sz; u++){
            I.a[u][u] = 1;
            for(int d = 0; d < 26; d++)
                if(!v[u] && !v[ch[u][d]])
                    M.a[u][ch[u][d]]++;
        }
    }
    ul A(int n){
        ul ans = 1;
        ul x = 26;
        while(n){
            if(n&1)    ans = ans*x;
            x = x*x;
            n /= 2;
        }
        return ans;                                                                    
    }
    mat add(mat x,mat y){
        mat z;
        for(int i = 0; i < sz; i++)
            for(int j = 0; j < sz; j++)
                z.a[i][j] = x.a[i][j]+y.a[i][j];
        return z;
    }
    mat calc(mat x,mat y){
        mat z;
        for(int i = 0; i < sz; i++)
            for(int j = 0; j < sz; j++){
                z.a[i][j] = 0;
                for(int k = 0; k < sz; k++)
                    z.a[i][j] += x.a[i][k]*y.a[k][j];
            }
        return z;
    }
    mat quick(int n){
        mat ans = I;
        mat x = M;
//        cout<<n<<endl;
        while(n){
            if(n&1)    ans = calc(ans,x);
            x = calc(x,x);
            n /= 2;
        }
        return ans;
    }
    ul S(int n){
        if(n == 1)    return 26;
        if(n&1)    return (1+A(n/2+1))*S(n/2)+A(n/2+1);
        else return (1+A(n/2))*S(n/2);
    }
    mat SUM(int n){
        if(n == 1)    return M;
        if(n&1)    return add(calc(add(I,quick(n/2+1)),SUM(n/2)),quick(n/2+1));
        else return calc(add(I,quick(n/2)),SUM(n/2));
    }
    ul getans(int n){
        ul ans = S(n);
    //    cout<<n<<endl;
    //    cout<<ans<<endl;
    //    cout<<sz<<endl;
        mat sum = SUM(n);
        for(int i = 0; i < sz; i++)
            ans -= sum.a[0][i];
    //    cout<<ans<<endl;
        return ans;
    }
}word;
char s[mx];
int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        word.init();
        for(int i = 1; i <= n; i++){
            scanf("%s",s);
            word.insert(s);
        }
        word.getfail();
        word.getmat();
        cout<<word.getans(m)<<endl;
    }
    return 0;
}