POJ 2778 DNA Sequence (AC自动机,矩阵乘法)

题意:给定n个不能出现的模式串，给定一个长度m，要求长度为m的合法串有多少种。

思路：用AC自动机，利用AC自动机上的节点做矩阵乘法。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<queue>
#define MOD 100000
char s[2005];
int ch(char x){
    if (x=='A') return 0;
    if (x=='C') return 1;
    if (x=='G') return 2;
    if (x=='T') return 3;
    return -1;
}
struct Trie{
    int fail[1005],next[1005][5],end[1005],L,root;
    int newnode(){
        for (int i=0;i<4;i++)
         next[L][i]=-1;
        end[L++]=-1;
        return L-1; 
    }
    void clear(){
         L=0;
         root=newnode();
    }
    void insert(char s[]){
        int len=strlen(s),now=root;
        for (int i=0;i<len;i++){
            if (next[now][ch(s[i])]==-1) next[now][ch(s[i])]=newnode();
            now=next[now][ch(s[i])];
        }
        end[now]=1;
    }
    void build(){
        std::queue<int>Q;
        int now;
        for (int i=0;i<4;i++)
         if (next[root][i]==-1)
          next[root][i]=root;
         else{
          fail[next[root][i]]=root;
          Q.push(next[root][i]);
         } 
        while (!Q.empty()){
            now=Q.front();Q.pop();
            if (end[fail[now]]==1) end[now]=1;
            for (int i=0;i<4;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]);
                }
            }
        } 
    }
}ac;
struct Matrix{
    int n,mx[105][105];
    Matrix(){}
    Matrix(int x){
        n=x;
        for (int i=0;i<n;i++)
         for (int j=0;j<n;j++)
          mx[i][j]=0;
    }    
};
Matrix operator *(Matrix a,Matrix b){
    int n=a.n;
    Matrix c=Matrix(n);
    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.mx[i][k]*b.mx[k][j]%MOD;
          c.mx[i][j]=(c.mx[i][j]+tmp)%MOD;
      }
    return c;   
} 
Matrix powm(Matrix t,int m){
    Matrix r=Matrix(t.n);
    for (int i=0;i<=t.n;i++)
     r.mx[i][i]=1;
    while (m){
        if (m%2) r=r*t;
        t=t*t;
        m/=2;
    }
    return r;
}
int main(){
    int m;
    int n;
    while (scanf("%d%d",&n,&m)!=EOF){
        ac.clear();
        for (int i=0;i<n;i++){
            scanf("%s",s);
            ac.insert(s);
        }
        ac.build();
        Matrix a=Matrix(ac.L);
        for (int i=0;i<ac.L;i++)
          for (int j=0;j<4;j++)
            if (ac.end[ac.next[i][j]]==-1)
                a.mx[i][ac.next[i][j]]++;
        a=powm(a,m);
        int ans=0;
        for (int i=0;i<a.n;i++)
         ans=(ans+a.mx[0][i])%MOD;
        printf("%d
",ans); 
    }
}