zoj 3822 Domination 概率DP

最新推荐文章于 2018-11-03 18:08:20 发布

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教育者埃德华购买了一块大小为N行M列的装饰棋盘，并每天随机放置一枚棋子。几天后，棋盘被棋子全部占据。埃德华想计算将空白棋盘变为完全占据状态所需的平均天数。

Domination

Time Limit: 8 Seconds Memory Limit: 131072 KB Special Judge

Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboard with N rows and M columns.

Every day after work, Edward will place a chess piece on a random empty cell. A few days later, he found the chessboard was dominated by the chess pieces. That means there is at least one chess piece in every row. Also, there is at least one chess piece in every column.

"That's interesting!" Edward said. He wants to know the expectation number of days to make an empty chessboard of N × M dominated. Please write a program to help him.

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:

There are only two integers N and M (1 <= N, M <= 50).

Output

For each test case, output the expectation number of days.

Any solution with a relative or absolute error of at most 10^-8 will be accepted.

Sample Input

2
1 3
2 2

Sample Output

3.000000000000
2.666666666667

Author: JIANG, Kai

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
#define N 55
#define M 2505

int n,m;
double dp[N][N][M];


int main()
{
    int _T;
    scanf("%d", &_T);
    while(_T--)
    {
        scanf("%d%d", &n, &m);
        memset(dp, 0, sizeof dp);
        dp[0][0][0] = 1.0;
        int tt = n * m;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                for(int k = 1; k <= i * j; k++)
                {
                    int tmp = tt - k + 1;
                    dp[i][j][k] += dp[i - 1][j - 1][k - 1] * (double)(n - i + 1) * (m - j + 1) ;
                    dp[i][j][k] += dp[i][j - 1][k - 1] * (double)i * (m - j + 1) ;
                    dp[i][j][k] += dp[i - 1][j][k - 1] * (double)(n - i + 1) * j ;
                    if(i != n || j != m)
                        dp[i][j][k] += dp[i][j][k - 1] * (double)(i * j - k + 1) ;
                    dp[i][j][k] /= tmp;
                    //printf("dp[%d][%d][%d] = %.10f\n", i, j, k, dp[i][j][k]);

                }
        double ans = 0.0;
        for(int i = 0; i <= tt; i++)
            ans += dp[n][m][i] * i;
        printf("%.10f\n", ans);
    }
    return 0;
}

/*

2
1 3
2 2

10
1 4
1 5
2 4
2 5
3 4
3 5
4 4

*/