POJ2778(SummerTrainingDay10-B AC自动机+矩阵快速幂)

DNA Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17160		Accepted: 6616

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

Source

POJ Monthly--2006.03.26,dodo

 

 //2017-08-10
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <queue>
 #define ll long long

 using namespace std;

 const int K = ;
 const int N = ;
 const int M = ;
 const int MOD = ;

 struct Matrix{
     ll a[M*M][M*M];
     int r, c;
 }mat, tmp;  

 Matrix multi(Matrix x, Matrix y)//矩阵乘法
 {
     Matrix z;
     memset(z.a, , sizeof(z.a));
     z.r = x.r, z.c = y.c;
     for(int i = ; i < x.r; i++){
         for(int k = ; k < x.c; k++)//加速优化
         {
             if(x.a[i][k] == ) continue;
             for(int j = ; j< y.c; j++)
                 z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD) % MOD;
         }
     }
     return z;
 }

 void Matrix_pow(int n)//矩阵快速幂
 {
     Matrix tmp;
     tmp.c = mat.c;
     tmp.r = mat.r;
     memset(tmp.a, , sizeof(tmp.a));
     for(int i = ; i < tmp.c; i++)
         tmp.a[i][i] = ;
     while(n){
         if(n & )
             tmp = multi(tmp, mat);
         mat = multi(mat, mat);
         n >>= ;
     }
     int ans = ;
     for(int i = ; i < tmp.c; i++)
         ans = (ans + tmp.a[][i]) % MOD;
     printf("%d\n", ans);
 }  

 struct AC_automation
 {
     //node nodes[N], *root, *superRoot, *cur;
     int nex[M*M][], fail[M*M], match[M*M];
     int root, CNT;
     int newNode(){
         for(int i = ; i < K; i++)
             nex[CNT][i] = -;
         match[CNT++] = ;
         return CNT-;
     }
     int Hash(char ch)
     {
         if(ch == 'A')return ;
         else if(ch == 'C')return ;
         else if(ch == 'T')return ;
         else if(ch == 'G')return ;
     }
     void init(){
         CNT = ;
         root = newNode();
     }
     void Insert(char s[]){//向字典树中插入一个字符串
         int n = strlen(s);
         int cur = root;
         for(int i = ; i < n; i++){
             int p = Hash(s[i]);
             if(nex[cur][p] == -)
                   nex[cur][p] = newNode();
             cur = nex[cur][p];
         }
         match[cur]++;
     }
     void build(){//构建自动机
         queue<int> que;
         fail[root] = root;
         for(int i = ; i < K; i++){
             if(nex[root][i] == -)
                 nex[root][i] = root;
             else{
                 fail[nex[root][i]] = root;
                 que.push(nex[root][i]);
             }
         }
         while(!que.empty()){
             int cur = que.front();
             if(match[fail[cur]])match[cur] = ;
             que.pop();
             for(int i = ; i < K; i++){
                 if(nex[cur][i] == -){
                     nex[cur][i] = nex[fail[cur]][i];
                 }else{
                     fail[nex[cur][i]] = nex[fail[cur]][i];
                     que.push(nex[cur][i]);
                 }
             }
         }
     }
     void to_marix(){
         memset(mat.a, , sizeof(mat.a));
         mat.r = mat.c = CNT;
         for(int i = ; i < CNT; i++){
             for(int j = ; j < ; j++)
                   if(!match[nex[i][j]])
                     mat.a[i][nex[i][j]]++;
         }
         // for(int i = 0; i < CNT; i++){
         //     for(int j = 0; j < CNT; j++)
         //           cout<<mat.a[i][j]<<" ";
         //     cout<<endl;
         // }
     }
 };

 char str[M];
 AC_automation ac;

 int main()
 {
     int n, m;
     while(scanf("%d%d", &m, &n)!=EOF)
     {
         ac.init();
         for(int i = ; i < m; i++){
             scanf("%s", str);
             ac.Insert(str);
         }
         ac.build();
         ac.to_marix();
         Matrix_pow(n);
     }

     return ;
 }