poj2778DNA Sequence (AC自動機+矩陣快速冪)

標籤：

轉載請註明出處： http://www.cnblogs.com/fraud/           ——by fraud

 

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 3ATACAGAA

Sample Output

36

Source

POJ Monthly--2006.03.26,dodo 題意構造一個長度為n的DNA序列，要求其中不得出現m個禁止的字串中的任意一個一道很明顯的矩陣快速冪的題。先通過AC自動機得出一個鄰接矩陣，然後快速冪。

  1 #include <iostream>  2 #include <cstring>  3 #include <cstdio>  4 #include <queue>  5 using namespace std;  6 #define REP(A,X) for(int A=0;A<X;A++)  7 #define MAXN    100010  8   9 int p[MAXN][4]; 10 int tail[MAXN]; 11 int fail[MAXN]; 12 int root,tot; 13 const long long MOD =100000; 14 struct Matrix{ 15     int n; 16     int mat[110][110]; 17     Matrix(){} 18     Matrix(int _n){ 19         n=_n; 20         REP(i,n) 21             REP(j,n)mat[i][j]=0; 22     } 23     void init() 24     { 25         REP(i,tot) 26         REP(j,tot)mat[i][j]=0; 27     } 28     void unit() 29     { 30         REP(i,tot) 31         REP(j,tot)mat[i][j]=i==j?1:0; 32     } 33     Matrix operator *(const Matrix &a)const { 34         Matrix ret(n); 35         REP(i,n) 36         REP(j,n) 37         REP(k,n) 38         { 39             int tmp=(long long)mat[i][k]*a.mat[k][j]%MOD; 40             ret.mat[i][j]=(ret.mat[i][j]+tmp)%MOD; 41         } 42         return ret; 43     } 44 }; 45 int newnode() 46 { 47     REP(i,4)p[tot][i]=-1; 48     tail[tot++]=0; 49     return tot-1; 50 } 51 void init() 52 { 53     tot=0; 54     root=newnode(); 55 } 56 int a[MAXN]; 57 void insert(char *s){ 58     int len=strlen(s); 59     REP(i,len) 60     { 61         if(s[i]==‘A‘)a[i]=0; 62         else if(s[i]==‘C‘)a[i]=1; 63         else if(s[i]==‘G‘)a[i]=2; 64         else if(s[i]==‘T‘)a[i]=3; 65     } 66     int now= root ; 67     REP(i,len) 68     { 69         if(p[now][a[i]]==-1)p[now][a[i]]=newnode(); 70         now=p[now][a[i]]; 71     } 72     tail[now]++; 73 } 74 void build() 75 { 76     int now=root; 77     queue<int >q; 78     fail[root]=root; 79     REP(i,4){ 80         if(p[root][i]==-1){ 81             p[root][i]=root; 82         } 83         else { 84             fail[p[root][i]]=root; 85             q.push(p[root][i]); 86         } 87     } 88     while(!q.empty()) 89     { 90         now =q.front(); 91         q.pop(); 92         if(tail[fail[now]])tail[now]=1; 93         REP(i,4){ 94             if(p[now][i]==-1){ 95                 p[now][i]=p[fail[now]][i]; 96             }else{ 97                 fail[p[now][i]]=p[fail[now]][i]; 98                 q.push(p[now][i]); 99             }100         }101     }102 }103 char s[MAXN];104 Matrix Mat;105 int main()106 {107     ios::sync_with_stdio(false);108     int n,m;109     while(cin>>m>>n){110         init();111         REP(i,m){112             cin>>s;113             insert(s);114         }115         build();116         Mat.n=tot;117         Mat.init();118         REP(i,tot){119             REP(j,4){120                 if(!tail[p[i][j]])Mat.mat[i][p[i][j]]++;121             }122         }123         Matrix tmp(tot);124         tmp.unit();125         while(n){126             if(n&1)tmp=tmp*Mat;127             Mat=Mat*Mat;128             n>>=1;129         }130         int ans=0;131         REP(i,tot)ans+=tmp.mat[0][i];132         ans%=MOD;133         cout<<ans<<endl;134 135     }136     return 0;137 }

代碼君 

poj2778DNA Sequence (AC自動機+矩陣快速冪)