poj2778

题意，给出一些串，问m个字符组成的串不包含这些串的有多少种情况。

自动机加矩阵快速幂。

矩阵相乘就是第i行第j列的数就是前一个矩阵的第i行，与后一矩阵的第j列，要保证前一个矩阵的列数等于后一个矩阵的行数。

构造的时候，只要把那些通向结点或通向的点的fild指向结点的节点去掉，其他正常

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<map>
using namespace std;
const int maxa = 105;
const int cha = 4;
#define LL long long
int n, m, k;
map<char, int> mp;
const int mod = 100000;
struct Matrix{
    int a[maxa][maxa];
    Matrix(){
        memset(a, 0, sizeof(a));
    }
    Matrix operator *(const Matrix &b)const
    {
        Matrix ret;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                for(int k=0;k<n;k++)
                {
                    int tmp=(long long)a[i][k]*b.a[k][j]%mod;
                    ret.a[i][j]=(ret.a[i][j]+tmp)%mod;
                }
        return ret;
    }/*
    Matrix operator *(const Matrix &b) const{
        Matrix tmp;
        for(int i =0 ;i < n; i++){;
            for(int j = 0; j < n; j++){
                for(int k = 0;k  <n; k++){
                    LL lo = (LL) (tmp.a[i][j]) + (LL)(a[i][k]) *b.a[k][j];
                    lo %= mod;
                    tmp.a[i][j] = lo;
                }
            }
        }
        return tmp;
    }*/
};
struct Tire{
    int next[maxa][cha], fail[maxa], end[maxa];
    int root, L;
    int newnode(){
        for(int i = 0; i < cha; 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++){
            int x = mp[buf[i]];
            if(next[now][x] == -1)
                next[now][x] = newnode();
            now = next[now][x];
            //printf("%d ", now);
        }//puts("");
        end[now] ++;
    }
    void build(){
        queue<int>Q;
        fail[root] = root;
        for(int i = 0; i < cha; 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();
            if(end[fail[now]]){
                end[now] = 1;
            }
            Q.pop();
           // end[now] |= end[fail[now]];
            for(int i = 0; i < cha; 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]);
                   // printf("**%d %d\n",next[now][i],next[fail[now]][i]);
                }
            }
        }
    }
    Matrix make_Maxtrix(){
        Matrix tmp;
        for(int i = 0; i < L; i++){
            for(int k = 0; k < cha; k++){
                if(!end[next[i][k]]){
                    tmp.a[i][next[i][k]] ++;
                }
            }
        }
        return tmp;
    }
};
Matrix pow(Matrix a,int m)
{
    Matrix ret;// = Matrix(a.n);
    for(int i = 0; i < n; i++)
        ret.a[i][i]=1;
    Matrix tmp=a;
    while(m)
    {
        if(m&1)ret=ret*tmp;
        tmp=tmp*tmp;
        m>>=1;
    }
    return ret;
}
char buf[1000005];
Tire ac;
int main(){

mp[‘A‘] = 0;
mp[‘T‘] = 1;
mp[‘G‘] = 2;
mp[‘C‘] = 3;
    int N, M;
    while(scanf("%d%d", &N, &M)!=EOF){
        ac.init();
        while(N--){
            scanf("%s", buf);
            ac.insert(buf);
        }
        ac.build();
        n =  ac.L;
        Matrix a = ac.make_Maxtrix();
        Matrix b = pow(a, M);;
        int ans =0 ;
        for(int i = 0; i < n ;i++){
            ans += b.a[0][i];
            ans %= mod;
        }printf("%d\n", ans);
    }
}
/*
abcdefg
bcdefg
cdef
de
e
ssaabcdefg

*/