poj 2278 DNASequnce AC自动机

地址：http://poj.org/problem?id=2778

题目：

DNA Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15453		Accepted: 5964

Description

It's well known that DNA Sequence is a sequence only contains A, C, T and G, and it's very useful to analyze a segment of DNA Sequence，For example, if a animal's DNA sequence contains segment ATC then it may mean that the animal may have a genetic disease. Until now scientists have found several those segments, the problem is how many kinds of DNA sequences of a species don't contain those segments.

Suppose that DNA sequences of a species is a sequence that consist of A, C, T and G，and the length of sequences is a given integer n.

Input

First line contains two integer m (0 <= m <= 10), n (1 <= n <=2000000000). Here, m is the number of genetic disease segment, and n is the length of sequences.

Next m lines each line contain a DNA genetic disease segment, and length of these segments is not larger than 10.

Output

An integer, the number of DNA sequences, mod 100000.

Sample Input

4 3
AT
AC
AG
AA

Sample Output

36

思路： 用病毒源建立ac自动机，然后所有节点可看做状态转移图。呃，懒得画图了，你们可以百度下。然后根据转移图建立矩阵，之后快速幂即可。
　　之前一直忘了把end【x】为1的节点的所有子节点的end标记为1，wa的不省人事。

 #include <queue>
 #include <cstring>
 #include <cstdio>
 using namespace std;

 const long long mod=1e5;
 struct MM
 {
     int r,c;
     long long mx[][];
     MM(int rr=,int cc=){r=rr,c=cc;}
     friend MM operator *(MM ta,MM tb)
     {
         MM tc(ta.r,tb.c);
         for(int i=;i<tc.r;i++)
             for(int j=;j<tc.c;j++)
             {
                 tc.mx[i][j]=;
                 for(int k=;k<tb.r;k++)
                     tc.mx[i][j]=(tc.mx[i][j]+ta.mx[i][k]*tb.mx[k][j])%mod;
             }
         return tc;
     }
     friend MM operator ^(MM ta,int num)
     {
         MM ret(ta.r,ta.c);                          //r==c
         memset(ret.mx,,sizeof ret.mx);
         for(int i=;i<ta.r;i++) ret.mx[i][i]=;    //µ¥Î»¾ØÕó
         while(num)
         {
             if(num&)
                 ret=ta*ret;
             num>>=;
             ta=ta*ta;
         }
         return ret;
     }
 }mm;
 struct AC_auto
 {
     const static int LetterSize = ;
     const static int TrieSize = *;

     int tot,root,fail[TrieSize],end[TrieSize],next[TrieSize][LetterSize];

     int newnode(void)
     {
         memset(next[tot],-,sizeof(next[tot]));
         end[tot] = ;
         return tot++;
     }

     void init(void)
     {
         tot = ;
         root = newnode();
     }

     int getidx(char x)
     {
         if(x=='A')return ;
         else if(x=='C')return ;
         else if(x=='G')return ;
         else return ;
     }

     void insert(char *ss)
     {
         int len = strlen(ss);
         int now = root;
         for(int i = ; i < len; i++)
         {
             int idx = getidx(ss[i]);
             if(next[now][idx] == -)
                 next[now][idx] = newnode();
             now = next[now][idx];
         }
         end[now]=;
     }

     void build(void)
     {
         queue<int>Q;
         fail[root] = root;
         for(int i = ; i < LetterSize; i++)
             if(next[root][i] == -)
                 next[root][i] = root;
             else
                 fail[next[root][i]] = root,Q.push(next[root][i]);
         while(Q.size())
         {
             int now = Q.front();Q.pop();
             if(end[fail[now]])end[now]=;
             for(int i = ; i < LetterSize; 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]);
         }
     }

     int buildmm(int n)
     {
         int cnt = ;
         mm.c=mm.r=tot;
         for(int i = ; i < tot; i++)
         for(int j = ; j < ; j++)
         if(!end[next[i][j]])
             mm.mx[i][next[i][j]]++;
         mm=mm ^ n;
         for(int i = ; i < tot; i++)
             cnt = (cnt + mm.mx[][i]) % mod;
         return cnt;
     }
 };

 AC_auto ac;
 char sa[];
 int main(void)
 {
     int n,m;
     scanf("%d%d",&n,&m);
     ac.init();
     for(int i=;i<=n;i++)
         scanf("%s",sa),ac.insert(sa);
     ac.build();
     printf("%d\n",ac.buildmm(m));
     return ;
 }