题目链接：zoj 3822 Domination

题目大意：给定一个N∗M的棋盘，每次任选一个位置放置一枚棋子，直到每行每列上都至少有一枚棋子，问放置棋子个数的期望。

解题思路：大白书上概率那一张有一道类似的题目，可是由于时间比較久了，还是略微想了一下。

dp[i][j][k]表示i行j列上均有至少一枚棋子，而且消耗k步的概率（k≤i∗j）,由于放置在i+1~n上等价与放在i+1行上，同理列也是如此。所以有转移方程：

• dp[i][j][k+1]+=dp[i][j][k]∗(n−k)(S−k)
• dp[i+1][j][k+1]+=dp[i][j][k]∗(N−i)∗j(S−k)
• dp[i][j+1][k+1]+=dp[i][j][k]∗(M−j)∗i(S−k)
• dp[i+1][j+1][k+1]+=dp[i][j][k]∗(N−i)∗(M−j)(S−k)

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn = 55;

const int maxm = 2505;

int N, M;

double dp[maxn][maxn][maxm];

double solve () {

    int S = N * M;

    memset(dp, 0, sizeof(dp));

    dp[1][1][1] = 1;

    for (int i = 1; i <= N; i++) {

        for (int j = 1; j <= M; j++) {

            int n = i * j;

            for (int k = max(i, j); k <= n; k++) {

                dp[i][j][k+1] += dp[i][j][k] * (n - k) / (S - k);

                dp[i+1][j][k+1] += dp[i][j][k] * (N - i) * j / (S - k);

                dp[i][j+1][k+1] += dp[i][j][k] * (M - j) * i / (S - k);

                dp[i+1][j+1][k+1] += dp[i][j][k] * (N - i) * (M - j) / (S - k);

            }

        }

    }

    /*

    for (int i = 1; i <= N; i++) {

        for (int j = 1; j <= M; j++) {

            printf("%d %d:", i, j);

            for (int k = max(i, j); k <= i * j; k++)

                printf("%.3lf ", dp[i][j][k]);

            printf("\n");

        }

    }

    */

    double ans = 0;

    for (int i = max(N, M); i <= S; i++)

        ans += (dp[N][M][i] - dp[N][M][i-1]) * i;

    return ans;

}

int main () {

    int cas;

    scanf("%d", &cas);

    while (scanf("%d%d", &N, &M) == 2) {

        printf("%.8lf\n", solve());

    }

    return 0;

}