POJ-2778 ac自动机+矩阵快速幂

DNA Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15422 Accepted: 5954

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

题目链接：

http://poj.org/problem?id=2778

题目大意：

       有m种DNA序列式有疾病的，问有多少种长度为n的DNA序列不包含任何一种有疾病的DNA序列。

解题思路：

        首先根据给出的序列构造ac自动机，自动机的每个节点是一种状态，在序列末尾添加一个字符就是一次状态转移。这样这道题就是要在每个序列末尾的节点做上标记，然后求在自动机上进行n次状态转移且不经过任何标记的方法数。

        我们将自动机看作一个有向图，问题转化为求从节点a到节点b经过n个节点的方法数。我们得到该有向图的邻接矩阵，求邻接矩阵的n次幂mat，mat[a][b]即为答案。

AC代码：

import java.util.*;public class Main {static int n,ct;static long m,mod=100000,ans;static int[] mp=new int[305];static int[] mm=new int[105];static int[] ff=new int[105];static int[][] ch=new int[105][4];static long[][] bb=new long[105][105];static long[][] xx=new long[105][105];static long[][] res=new long[105][105];static Queue<Integer> q=new LinkedList<Integer>();static void clear(int u){ch[u][0]=ch[u][1]=ch[u][2]=ch[u][3]=0;}static void insert(char[] s){int u=0,a;for(int i=0;i<s.length;i++){a=mp[s[i]];if(ch[u][a]==0){ch[u][a]=++ct;clear(ct);}u=ch[u][a];}mm[u]=1;}static void build(){q.clear();int u,v,r;for(int i=0;i<4;i++){u=ch[0][i];if(u>0) { q.add(u);ff[u]=0;}}while(!q.isEmpty()){r=q.poll();v=ff[r];if(mm[v]==1) mm[r]=1;for(int i=0;i<4;i++){u=ch[r][i];if(u==0) ch[r][i]=ch[v][i];else { ff[u]=ch[v][i];q.add(u);}}}}static void get_mat(){for(int i=0;i<=ct;i++){for(int j=0;j<=ct;j++)xx[i][j]=0;for(int j=0;j<4;j++)if(mm[ch[i][j]]==0) xx[i][ch[i][j]]++;}}static void mul_mat(long[][] xx,long[][] yy){for(int i=0;i<=ct;i++)for(int j=0;j<=ct;j++){bb[i][j]=0;for(int k=0;k<=ct;k++)bb[i][j]=(bb[i][j]+xx[i][k]*yy[k][j]%mod)%mod;}for(int i=0;i<=ct;i++)for(int j=0;j<=ct;j++)xx[i][j]=bb[i][j];}static void pow_mat(long n){for(int i=0;i<=ct;i++){for(int j=0;j<=ct;j++)res[i][j]=0;res[i][i]=1;}while(n>0){if((n&1)>0) mul_mat(res,xx);mul_mat(xx,xx);n>>=1;}}public static void main(String[] args) {Scanner in=new Scanner(System.in);mp['A']=0;mp['T']=1;mp['C']=2;mp['G']=3;while(in.hasNext()){n=in.nextInt();m=in.nextLong();clear(0);ct=0;Arrays.fill(mm,0);for(int i=1;i<=n;i++)insert(in.next().toCharArray());build();get_mat();pow_mat(m);ans=0;for(int i=0;i<=ct;i++)ans=(ans+res[0][i])%mod;System.out.println(ans);}}}