HDOJ 5411 CRB and Puzzle 矩阵高速幂

直接构造矩阵,最上面一行加一排1.高速幂计算矩阵的m次方,统计第一行的和

CRB and Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 133    Accepted Submission(s): 63

Problem Description

CRB is now playing Jigsaw Puzzle.
There are N kinds of pieces with infinite supply.
He can assemble one piece to the right side of the previously assembled one.
For each kind of pieces, only restricted kinds can be assembled with.
How many different patterns he can assemble with at most M pieces?

(Two patterns P and Q are considered different if their lengths are different or there exists an integer j such that j-th piece of P is different from corresponding piece of Q.)

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains two integers N, M denoting the number of kinds of pieces and the maximum number of moves.
Then N lines follow. i-th line is described as following format.
k a1 a2 ... ak
Here k is the number of kinds which can be assembled to the right of the i-th kind. Next k integers represent each of them.
1 ≤ T ≤ 20
1 ≤ N ≤ 50
1 ≤ M ≤ 105
0 ≤ k ≤ N
1 ≤ a1 < a2 < … < ak ≤ N

 

Output

For each test case, output a single integer - number of different patterns modulo 2015.

 

Sample Input

1
3 2
1 2
1 3
0

 

Sample Output

6
Hint
possible patterns are ∅, 1, 2, 3, 1→2, 2→3
 

 

Author

KUT（DPRK）

 

Source

2015 Multi-University Training Contest 10

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月20日 星期四 23时25分19秒
File Name     :HDOJ5411.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int mod=2015;

int n,m;

struct Matrix
{
	int m[60][60];
	Matrix() { memset(m,0,sizeof(m)); }
	void getE()
	{
		for(int i=0;i<n;i++) m[i][i]=1;
	}
	void toString()
	{
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<n;j++)
			{
				printf("%d,",m[i][j]);
			}
			putchar(10);
		}
	}
};

Matrix Mulit(Matrix a,Matrix b)
{
	Matrix M;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<n;j++)
		{
			int temp=0;
			for(int k=0;k<n;k++)
			{
				temp=(temp+a.m[i][k]*b.m[k][j])%mod;
			}
			M.m[i][j]=temp;
		}
	}
	return M;
}

Matrix QuickPow(Matrix a,int x)
{
	Matrix e;
	e.getE();
	while(x)
	{
		if(x&1) e=Mulit(e,a);
		a=Mulit(a,a);
		x/=2;
	}
	return e;
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d%d",&n,&m);
		Matrix M;
		for(int i=1;i<=n;i++)
		{
			int k,x;
			scanf("%d",&k);
			for(int j=0;j<k;j++)
			{
				scanf("%d",&x);
				M.m[i][x]=1;
			}
		}
		n++;
		for(int i=0;i<n;i++) M.m[0][i]=1;
		Matrix mt=QuickPow(M,m);
		int ans=0;
		for(int i=0;i<n;i++)
		{
			ans=(ans+mt.m[0][i])%mod;
		}
		printf("%d\n",ans);
	}
    
    return 0;
}