HDU 5411 CRB and puzzle (Dp + 矩阵快速幂）

CRB and Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory
Limit: 65536/65536 K (Java/Others)

Total Submission(s): 483    Accepted Submission(s): 198


Problem Description

CRB is now playing Jigsaw Puzzle.

There are 

 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 

 pieces?
(Two patterns 

 and 

 are
considered different if their lengths are different or there exists an integer 

 such
that 

-th
piece of 

 is
different from corresponding piece of 

.)

 

Input

There are multiple test cases. The first line of input contains an integer 

,
indicating the number of test cases. For each test case:

The first line contains two integers 

, 

 denoting
the number of kinds of pieces and the maximum number of moves.

Then 

 lines
follow. 

-th
line is described as following format.

k 
























Here 

 is
the number of kinds which can be assembled to the right of the 

-th
kind. Next 

 integers
represent each of them.

1 ≤ 

 ≤
20

1 ≤ 

 ≤
50

1 ≤ 

 ≤ 






0 ≤ 

 ≤ 


1 ≤ 



 < 



 <
… < 



 ≤
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
Hintpossible patterns are ∅, 1, 2, 3, 1→2, 2→3

 

Author

KUT（DPRK）

解题思路：

DP方程很容易想到 dp[i][j] = sum(dp[i-1][k] <k,j>连通） 构造矩阵用矩阵快速幂加速即可。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#define LL long long
using namespace std;
const int MAXN = 55 + 10;
const int mod = 2015;
int n, m;
struct Matrix
{
int m[MAXN][MAXN];
Matrix(){memset(m, 0, sizeof(m));}
Matrix operator * (const Matrix &b)const
{
Matrix res;
for(int i=1;i<=n+1;i++)
{
for(int j=1;j<=n+1;j++)
{
for(int k=1;k<=n+1;k++)
{
res.m[i][j] = (res.m[i][j] + m[i][k] * b.m[k][j]) % mod;
}
}
}
return res;
}
};
Matrix pow_mod(Matrix a, int b)
{
Matrix res;
for(int i=1;i<=n+1;i++) res.m[i][i] = 1;
while(b)
{
if(b & 1) res = res * a;
a = a * a;
b >>= 1;
}
return res;
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
Matrix a, b;
scanf("%d%d", &n, &m);
for(int i=1;i<=n+1;i++) a.m[i][n+1] = 1;
for(int i=1;i<=n;i++)
{
int x, k;scanf("%d", &k);
for(;k--;)
{
scanf("%d", &x);
a.m[i][x] = 1;
}
}
a = pow_mod(a, m);
int ans = 0;
for(int i=1;i<=n+1;i++) ans = (ans + a.m[i][n+1]) % mod;
printf("%d\n", ans);
}
return 0;
}