poj2778DNA Sequence (AC自动机+矩阵快速幂)

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DNA Sequence

Time Limit: 1000MS

Memory Limit: 65536K

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

 

题意

构造一个长度为n的DNA序列，要求其中不得出现m个禁止的字符串中的任意一个

一道很明显的矩阵快速幂的题。先通过AC自动机得出一个邻接矩阵，然后快速幂。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <queue>
 using namespace std;
 #define REP(A,X) for(int A=0;A<X;A++)
 #define MAXN    100010

 int p[MAXN][];
 int tail[MAXN];
 int fail[MAXN];
 int root,tot;
 const long long MOD =;
 struct Matrix{
     int n;
     int mat[][];
     Matrix(){}
     Matrix(int _n){
         n=_n;
         REP(i,n)
             REP(j,n)mat[i][j]=;
     }
     void init()
     {
         REP(i,tot)
         REP(j,tot)mat[i][j]=;
     }
     void unit()
     {
         REP(i,tot)
         REP(j,tot)mat[i][j]=i==j?:;
     }
     Matrix operator *(const Matrix &a)const {
         Matrix ret(n);
         REP(i,n)
         REP(j,n)
         REP(k,n)
         {
             int tmp=(long long)mat[i][k]*a.mat[k][j]%MOD;
             ret.mat[i][j]=(ret.mat[i][j]+tmp)%MOD;
         }
         return ret;
     }
 };
 int newnode()
 {
     REP(i,)p[tot][i]=-;
     tail[tot++]=;
     return tot-;
 }
 void init()
 {
     tot=;
     root=newnode();
 }
 int a[MAXN];
 void insert(char *s){
     int len=strlen(s);
     REP(i,len)
     {
         if(s[i]=='A')a[i]=;
         else if(s[i]=='C')a[i]=;
         else if(s[i]=='G')a[i]=;
         else if(s[i]=='T')a[i]=;
     }
     int now= root ;
     REP(i,len)
     {
         if(p[now][a[i]]==-)p[now][a[i]]=newnode();
         now=p[now][a[i]];
     }
     tail[now]++;
 }
 void build()
 {
     int now=root;
     queue<int >q;
     fail[root]=root;
     REP(i,){
         if(p[root][i]==-){
             p[root][i]=root;
         }
         else {
             fail[p[root][i]]=root;
             q.push(p[root][i]);
         }
     }
     while(!q.empty())
     {
         now =q.front();
         q.pop();
         if(tail[fail[now]])tail[now]=;
         REP(i,){
             if(p[now][i]==-){
                 p[now][i]=p[fail[now]][i];
             }else{
                 fail[p[now][i]]=p[fail[now]][i];
                 q.push(p[now][i]);
             }
         }
     }
 }
 char s[MAXN];
 Matrix Mat;
 int main()
 {
     ios::sync_with_stdio(false);
     int n,m;
     while(cin>>m>>n){
         init();
         REP(i,m){
             cin>>s;
             insert(s);
         }
         build();
         Mat.n=tot;
         Mat.init();
         REP(i,tot){
             REP(j,){
                 if(!tail[p[i][j]])Mat.mat[i][p[i][j]]++;
             }
         }
         Matrix tmp(tot);
         tmp.unit();
         while(n){
             if(n&)tmp=tmp*Mat;
             Mat=Mat*Mat;
             n>>=;
         }
         int ans=;
         REP(i,tot)ans+=tmp.mat[][i];
         ans%=MOD;
         cout<<ans<<endl;

     }
     return ;
 }

代码君