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HDOJ 5411 CRB and Puzzle 矩陣快速冪

編輯：關於C++

直接構造矩陣,最上面一行加一排1.快速冪計算矩陣的m次方,統計第一行的和

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

Sample Output

6
Hintpossible 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時25分19秒
File Name     :HDOJ5411.cpp
************************************************ */

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include

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