hdu5411 CRB and Puzzle[矩阵优化dp]

 

1|0CRB and Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1177    Accepted Submission(s): 468

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 a_{1}\ a_{2}\ ...\ a_{k}
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 ≤ 10^5
0 ≤ k ≤ N
1 ≤ a_{1} < a_{2} < … < a_{k} ≤ 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

 

 

Recommend

wange2014

 

同样的题：不同的写法，却得不出相同的答案。poj3233 Matrix Power Series

望大佬给蒟蒻答疑

 Select Code

#include<cstdio>
#include<cstring>
using namespace std;
const int mod=2015;
struct matrix{
	int s[102][102];
	matrix(){
		memset(s,0,sizeof s);
	}
}A,F;int n,m,T;int ans;
matrix operator *(const matrix &a,const matrix &b){
	matrix c;
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			for(int k=0;k<n;k++){
				c.s[i][j]+=a.s[i][k]*b.s[k][j];
			}
			c.s[i][j]%=mod;
		}
	}
	return c;
}
matrix fpow(matrix a,int p){
	matrix res;
	for(int i=0;i<n;i++) res.s[i][i]=1;
	for(;p;p>>=1,a=a*a) if(p&1) res=res*a;
	return res;
}
int main(){
	for(scanf("%d",&T);T--;){
		scanf("%d%d",&n,&m);
		matrix A;
		for(int i=0,k,x;i<n;i++){
			scanf("%d",&k);
			while(k--){
				scanf("%d",&x);x--;
				A.s[i][x]=1;
			}
		}
		for(int i=0;i<n;i++) A.s[i][i+n]=A.s[i+n][i+n]=1;
		n<<=1;
		A=fpow(A,m);
		n>>=1;
		ans=1;
		for(int i=0;i<n;i++){
        	for(int j=n;j<2*n;j++){
        		ans+=A.s[i][j];
			}
		}
		ans%=mod;
		printf("%d\n",ans);
	}
	return 0;
}

__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/6726629.html
关于博主：评论和私信会在第一时间回复。或者直接私信我。
版权声明：本博客所有文章除特别声明外，均采用 BY-NC-SA 许可协议。转载请注明出处！
声援博主：如果您觉得文章对您有帮助，可以点击文章右下角【推荐】一下。您的鼓励是博主的最大动力！