[HDU 5411 2015 Multi-University Training Contest 10F][매트릭스 쾌속 멱 에 한 줄 구조 법 추가]CRB and Puzzle 매트릭스 의 1 차방 에서 n 차방 의 수치 와

CRB and Puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 877    Accepted Submission(s): 351
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
 

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<functional>
#include<string>
#include<algorithm>
#include<time.h>
#include<bitset>
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
typedef int Int;
template <class T> inline void gmax(T &a,T b){if(b>a)a=b;}
template <class T> inline void gmin(T &a,T b){if(b<a)a=b;}
using namespace std;
const int N=51,M=0,L=0,Z=2015,maxint=2147483647,ms31=522133279,ms63=1061109567,ms127=2139062143;
const double eps=1e-8,PI=acos(-1.0);//.0
map<int,int>mop;
struct A{};
int casenum,casei;
int n,m,nn;
struct MX
{
	int v[N][N];
	void O(){MS(v,0);}//     
	void E(){MS(v,0);for(int i=0;i<nn;i++)v[i][i]=1;}//      
	MX operator * (const MX &b) //const   const       ——       b  
	{
		MX c;c.O();
		for(int k=0;k<nn;k++)
		{
			for(int i=0;i<nn;i++)
			{
				for(int j=0;j<nn;j++)
				{
					c.v[i][j]=(c.v[i][j]+v[i][k]*b.v[k][j]);
				}
			}
		}
		for(int i=0;i<nn;i++)
		{
			for(int j=0;j<nn;j++)c.v[i][j]%=Z;
		}
		/*
		              。
		v[i,j]      2015*2015*100   ，   4e8，   int
		     AC 。
		*/
		return c;
	}
	MX operator ^ (int p)
	{
		MX x;x.E();//       
		MX y;MC(y.v,v);
		while(p)
		{
			if(p&1)x=x*y;
			p/=2;
			y=y*y;
		}
		return x;
	}
}a;
int main()
{
	//fre();
	scanf("%d",&casenum);
	for(casei=1;casei<=casenum;casei++)
	{
		a.O();
		scanf("%d%d",&n,&m);
		for(int i=0;i<n;i++)
		{
			int k,x;scanf("%d",&k);
			for(int j=0;j<k;j++)
			{
				scanf("%d",&x);
				a.v[i][x-1]=1;
			}
		}
		for(int i=0;i<=n;i++)a.v[i][n]=1;
		nn=n+1;a=a^(m-1);//  ^m    E+A+A^2+...+A^(m-1)
		int ans=0;
		for(int i=0;i<nn;i++)
		{
			for(int j=0;j<nn;j++)ans+=a.v[i][j];
		}
		printf("%d
",ans%Z);
	}
	return 0;
}
/*
【  】
 n（1<=n<=50）   ，     。
            ，          。
     ，  0~m（1<=m<=1e5）             。

【  】
     

【  】
               。
aij    i          j
               ~

【     &&  】
     ，        。
 A^0+A^1+A^2+...+A^m，         ——
AE
OE
         ^(m+1)，      ——
AB
CD
B   A^0+A^1+A^2+...+A^m

             8 ，     TLE
             ，   TLE
           ，     AC，     >_<

===========★        ★============
        5  ！    ！
       ？
     
AB
CD
        
AB0
CD0
111
  m=1，    a^0，     ，        n+1 
  m=2，    a^1，      ，    
AB0 AB0
CD0 CD0
111 111
  m=3，         
    n*n               ，        A^2
      111……              
          
AB0
CD0
XX1   ，    。         AC 。

*/