Domination（概率DP）

Domination

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

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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

 

     题意：

     给出 T 组样例，每组样例有 N，M 两个数，代表 N 行 M 列。每次往棋盘放一只棋子，每个棋子可以覆盖所在行与列，输出放多少个棋子能覆盖整个棋盘的数学期望。

 

     思路：

     概率 DP。dp [ i ] [ j ] [ k ] 代表用 k 只棋子覆盖了 i 行，j 列（任意行，任意列，代表的是有 i 行，有 j 列而不是特指某 i 行，某 j 列）。转移方程的时候注意要分开讨论，因为如果当 i == n && j == m 的时候，如果之前已经是覆盖了的就不需要加上 dp [ i ] [ j ] [ k - 1 ] * （i * j - （k - 1）） / ( n * m - k + 1 ) 的概率了。

 

      AC：

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

using namespace std;

const int N = 55;
const int MAX = N * N;

double dp[N][N][MAX];

int main() {

    int t;
    scanf("%d", &t);

    while (t--) {
        int n, m;
        scanf("%d%d", &n, &m);

        double sum = n * m;

        memset(dp, 0, sizeof(dp));
        dp[0][0][0] = 1.0;

        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                for (int k = 1; k <= sum; ++k) {

                    double left = sum - (k - 1);
                    double p1 = (i * j - (k - 1)) / left;
                    double p2 = (n - (i - 1)) * j / left;
                    double p3 = (m - (j - 1)) * i / left;
                    double p4 = (n - (i - 1)) * (m - (j - 1)) / left;

                    if (i == n && j == m) {
                        dp[i][j][k] += dp[i - 1][j][k - 1] * p2;
                        dp[i][j][k] += dp[i][j - 1][k - 1] * p3;
                        dp[i][j][k] += dp[i - 1][j - 1][k - 1] * p4;
                    } else {
                        dp[i][j][k] += dp[i][j][k - 1] * p1;
                        dp[i][j][k] += dp[i - 1][j][k - 1] * p2;
                        dp[i][j][k] += dp[i][j - 1][k - 1] * p3;
                        dp[i][j][k] += dp[i - 1][j - 1][k - 1] * p4;
                    }
                }
            }
        }

        double res = 0;
        for (int i = 0; i <= sum; ++i) {
            res += dp[n][m][i] * i;
        }

        printf("%.12lf\n", res);
    }

    return 0;
}