POJ2778-DNA Sequence

DNA Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18029		Accepted: 6937

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：给出患病的DNA序列，问序列长度为n的，且不包含患病的DNA序列有多少种

解题思路：ac自动机+矩阵快速幂

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cctype>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

char ch[15];
int n, m;

char get(char ch)
{
	if (ch == 'A') return 0;
	if (ch == 'C') return 1;
	if (ch == 'G') return 2;
	else return 3;
}

struct Matrix
{
	int v[110][110];
	Matrix()
	{
		memset(v, 0, sizeof v);
	}
} dan;

Matrix mul(Matrix a, Matrix b, int d)
{
	Matrix ans;
	for (int i = 0; i < d; i++)
		for (int j = 0; j < d; j++)
			for (int k = 0; k < d; k++)
				(ans.v[i][j] += (1LL * a.v[i][k] * b.v[k][j]) % 100000) %= 100000;
	return ans;
}

Matrix pow(Matrix a, int k, int d)
{
	Matrix ans = dan;
	while (k)
	{
		if (k & 1) ans = mul(ans, a, d);
		k >>= 1;
		a = mul(a, a, d);
	}
	return ans;
}

struct Trie
{
	int next[110][4], fail[110], flag[110];
	int root, tot;
	int newnode()
	{
		for (int i = 0; i < 4; i++) next[tot][i] = -1;
		flag[tot++] = 0;
		return tot - 1;
	}
	void init()
	{
		tot = 0;
		root = newnode();
	}
	void insert(char ch[])
	{
		int k = root;
		for (int i = 0; ch[i]; i++)
		{
			if (next[k][get(ch[i])] == -1) next[k][get(ch[i])] = newnode();
			k = next[k][get(ch[i])];
		}
		flag[k] = 1;
	}
	void build()
	{
		queue<int>q;
		fail[root] = root;
		for (int i = 0; i < 4; i++)
		{
			if (next[root][i] == -1) next[root][i] = root;
			else
			{
				fail[next[root][i]] = root;
				q.push(next[root][i]);
			}
		}
		while (!q.empty())
		{
			int pre = q.front();
			q.pop();
			if (flag[fail[pre]]) flag[pre] = 1;
			for (int i = 0; i < 4; i++)
			{
				if (next[pre][i] == -1) next[pre][i] = next[fail[pre]][i];
				else
				{
					fail[next[pre][i]] = next[fail[pre]][i];
					q.push(next[pre][i]);
				}
			}
		}
	}
	void solve()
	{
		Matrix ans;
		for (int i = 0; i < tot; i++)
			for (int j = 0; j < 4; j++)
				if (!flag[next[i][j]]) ans.v[i][next[i][j]]++;
		for (int i = 0; i < tot; i++) dan.v[i][i] = 1;
		ans = pow(ans, m, tot);
		int sum = 0;
		for (int i = 0; i < tot; i++) (sum += ans.v[0][i]) %= 100000;
		printf("%d\n", sum);
	}
	void debug()
	{
		for (int i = 0; i < tot; i++)
		{
			printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], flag[i]);
			for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
			printf("]\n");
		}
	}
}ac;

int main()
{
	while (~scanf("%d %d", &n, &m))
	{
		ac.init();
		for (int i = 1; i <= n; i++)
		{
			scanf("%s", ch);
			ac.insert(ch);
		}
		ac.build();
		ac.solve();
	}
	return 0;
}