HDU 5411 CRB and puzzle (Dp + 矩阵高速幂）

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

 

Author

KUT（DPRK）

解题思路：
DP方程非常easy想到 dp[i][j] = sum(dp[i-1][k] <k,j>连通） 构造矩阵用矩阵高速幂加速就可以。

#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;
}