POJ-2778 ac自动机+矩阵快速幂-优快云博客

DNA Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total 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 3
AT
AC
AG
AA

Sample Output

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);
		}
	}
}