线性代数（矩阵乘法）：POJ 2778 DNA Sequence

DNA Sequence

 

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
　　
思路是这样的：把所有病毒片段放入AC自动机中，建立fail数组。如果一个状态的fail为病毒节点，则他自己也为病毒节点。最后按边建矩阵，快速幂。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <queue>
 using namespace std;
 const int maxn=;
 const int mod=;
 typedef unsigned long long ull;
 struct Matrix{
     int n;
     ull mat[maxn][maxn];
     Matrix(int n_,int on=){
         n=n_;memset(mat,,sizeof(mat));
         if(on)for(int i=;i<=n;i++)mat[i][i]=;
     }
     Matrix operator *(Matrix a){
         Matrix ret(n);
         unsigned long long l;
         for(int i=;i<=n;i++)
             for(int k=;k<=n;k++){
                 l=mat[i][k];
                 for(int j=;j<=n;j++)
                     (ret.mat[i][j]+=l*a.mat[k][j]%mod)%=mod;
             }
         return ret;
     }
     Matrix operator ^(long long k){
         Matrix ret(n,);
         while(k){
             if(k&)
                 ret=ret**this;
             k>>=;
             *this=*this**this;
         }
         return ret;
     }
 };

 struct AC_automation{
     bool tag[maxn];
     int cnt,rt,ch[maxn][],fail[maxn];
     AC_automation(){
         memset(tag,,sizeof(tag));
         memset(fail,,sizeof(fail));
         memset(ch,,sizeof(ch));cnt=rt=;
     }

     int ID(char c){
         if(c=='A')return ;
         else if(c=='C')return ;
         else if(c=='G')return ;
         else return ;
     }

     void Insert(char *s){
         int len=strlen(s),p=rt;
         for(int i=;i<len;i++)
             if(ch[p][ID(s[i])])
                 p=ch[p][ID(s[i])];
             else
                 p=ch[p][ID(s[i])]=++cnt;
         tag[p]=true;
     }

     void Build(){
         queue<int>q;
         for(int i=;i<;i++)
             if(ch[rt][i])
                 fail[ch[rt][i]]=rt,q.push(ch[rt][i]);
             else
                 ch[rt][i]=rt;

         while(!q.empty()){
             int x=q.front();q.pop();
             for(int i=;i<;i++)
                 if(ch[x][i]){
                     fail[ch[x][i]]=ch[fail[x]][i];
                     tag[ch[x][i]]|=tag[fail[ch[x][i]]];
                     q.push(ch[x][i]);
                 }
                 else
                     ch[x][i]=ch[fail[x]][i];
         }
     }

     void Solve(int k){
         Matrix A(cnt);
         for(int i=;i<=cnt;i++)
             for(int j=;j<;j++)
                 if(!tag[i]&&!tag[ch[i][j]])
                     A.mat[ch[i][j]][i]+=;
         A=A^k;
         long long ans=;
         for(int i=;i<=cnt;i++)
             ans+=A.mat[i][];
         printf("%lld\n",ans%mod);
     }
 }ac;
 char s[maxn];

 int main(){
 #ifndef ONLINE_JUDGE
     //freopen("","r",stdin);
     //freopen("","w",stdout);
 #endif
     int tot,n;
     scanf("%d%d",&tot,&n);
     while(tot--){
         scanf("%s",s);
         ac.Insert(s);
     }
     ac.Build();
     ac.Solve(n);
     return ;
 }