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[];
 int ch(char x){
     if (x=='A') return ;
     if (x=='C') return ;
     if (x=='G') return ;
     if (x=='T') return ;
     return -;
 }
 struct Trie{
     int fail[],next[][],end[],L,root;
     int newnode(){
         for (int i=;i<;i++)
          next[L][i]=-;
         end[L++]=-;
         return L-;
     }
     void clear(){
          L=;
          root=newnode();
     }
     void insert(char s[]){
         int len=strlen(s),now=root;
         for (int i=;i<len;i++){
             if (next[now][ch(s[i])]==-) next[now][ch(s[i])]=newnode();
             now=next[now][ch(s[i])];
         }
         end[now]=;
     }
     void build(){
         std::queue<int>Q;
         int now;
         for (int i=;i<;i++)
          if (next[root][i]==-)
           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]]==) end[now]=;
             for (int i=;i<;i++){
                 if (next[now][i]==-) 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[][];
     Matrix(){}
     Matrix(int x){
         n=x;
         for (int i=;i<n;i++)
          for (int j=;j<n;j++)
           mx[i][j]=;
     }
 };
 Matrix operator *(Matrix a,Matrix b){
     int n=a.n;
     Matrix c=Matrix(n);
     for (int i=;i<n;i++)
      for (int j=;j<n;j++)
       for (int k=;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=;i<=t.n;i++)
      r.mx[i][i]=;
     while (m){
         if (m%) r=r*t;
         t=t*t;
         m/=;
     }
     return r;
 }
 int main(){
     int m;
     int n;
     while (scanf("%d%d",&n,&m)!=EOF){
         ac.clear();
         for (int i=;i<n;i++){
             scanf("%s",s);
             ac.insert(s);
         }
         ac.build();
         Matrix a=Matrix(ac.L);
         for (int i=;i<ac.L;i++)
           for (int j=;j<;j++)
             if (ac.end[ac.next[i][j]]==-)
                 a.mx[i][ac.next[i][j]]++;
         a=powm(a,m);
         int ans=;
         for (int i=;i<a.n;i++)
          ans=(ans+a.mx[][i])%MOD;
         printf("%d\n",ans);
     }
 }