POJ 2778 (AC自动机+矩阵乘法)

POJ 2778 DNA Sequence

Problem : 给m个只含有（A,G,C,T）的模式串(m <= 10, len <=10), 询问所有长度为n的只含有(A,G,C,T)的串中有多少个不含有模式串的串。(n<=2000000000)
Solution :首先对所有模式串建立AC自动机。然后dp[i][j]表示长度为i，走到AC自动机的节点j这样的字符串满足条件的个数有多少，用AC自动机的边写出状态转移方程然后用矩阵快速幂加速运算。

#include <iostream>
#include <string>
#include <queue>

using namespace std;

const int N = 208;
const int mo = 100000;

int id[128];

struct Matrix
{
	int n;
	int a[N][N];
	Matrix(int n_, int p)
	{
		n = n_;
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < n; ++j)
			{
				a[i][j] = 0;
				if (i == j) a[i][j] = p;
			}
	}
	friend Matrix operator *(Matrix A, Matrix B)
	{
		Matrix C(A.n, 0);
		for (int i = 0; i < A.n; ++i)
			for (int j = 0; j < A.n; ++j)
				for (int k = 0; k < A.n; k++)
					C.a[i][j] = (C.a[i][j] + 1ll * A.a[i][k] * B.a[k][j] % mo) % mo; 
		return C;
	}
	void print()
	{
		for (int i = 0; i < n; ++i)
		{
			for (int j = 0; j < n; ++j) cout << a[i][j] << " ";
			cout << endl;
		}
	}
};

struct AC_Automan
{
	int next[N][4];
	int fail[N];
	int cnt[N];
	int root, tot;
	
	int newnode()
	{
		for (int i = 0; i <= 3; ++i) next[tot][i] = -1;
		fail[tot] = cnt[tot] = -1;
		return tot++;
	}
	void clear()
	{
		tot = 0;
		root = newnode();
	}
	void insert(const string &s)
	{
		int p = root;
		for (int i = 0, len = s.length(); i < len; ++i)
		{
			if (next[p][id[s[i]]] == -1) next[p][id[s[i]]] = newnode();
			p = next[p][id[s[i]]];
		}
		cnt[p] = 1;
	}
	void build()
	{
		queue <int> Q;
		Q.push(root);
		while (!Q.empty())
		{
			int p = Q.front(); Q.pop();
			for (int i = 0; i < 4; ++i)
			{
				if (~next[p][i])
				{
					if (p == root) fail[next[p][i]] = root; 
					else fail[next[p][i]] = next[fail[p]][i];
					Q.push(next[p][i]);
				}		
				else
				{
					if (p == root) next[p][i] = root;
					else next[p][i] = next[fail[p]][i];
				}
			}
		}
	}
	Matrix power(Matrix A, int y)
	{
		Matrix B(tot, 1);
		while (y)
		{
			if (y & 1) B = B * A;
			A = A * A;
			y >>= 1;
		}
		return B;
	}

	void solve(int num)
	{
		Matrix A(tot, 0);
		for (int i = 0; i <= tot; ++i)
		{
			for (int j = 0; j < 4; ++j)
			{
				int flag = 1;
				for (int temp = next[i][j]; temp != root; temp = fail[temp])
				{
					if (~cnt[temp]) flag = 0;
				}
				A.a[i][next[i][j]] += flag;
			}
		}
		A = power(A, num);
		int ans = 0;
		for (int i = 0; i < tot; ++i) ans = (ans + A.a[0][i]) % mo;
		cout << ans << endl;
	}
}ac;



int main()
{
	cin.sync_with_stdio(0);
	id['A'] = 0; id['G'] = 1; id['C'] = 2; id['T'] = 3;
	int m, n;
	while (cin >> m >> n)
	{
		ac.clear();
		for (int i = 1; i <= m; ++i)
		{
			string s; cin >> s;
			ac.insert(s);
		}
		ac.build();
		ac.solve(n);
	}
}