ZOJ3822——概率DP——Domination

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3822

/*
dp[i][j][k]定义为覆盖了i行j列用了>=k个棋子的概率
状态转移方程
dp[i][j][k] = dp[i-1][j][k-1]*1.0*(j*(n-i+1))/(n*m-k+1)
            + dp[i][j-1][k-1]*1.0*i*(m-j+1)/(n*m-k+1)
            + dp[i-1][j-1][k-1]*1.0*(n-i+1)*(m-j+1)/(n*m-k+1)
            + dp[i][j][k-1]*(1.0*i*j-k+1)/(n*m-k+1);//k-1已经被覆盖不能选
n-m-k+1表示k-1空格已经被覆盖，要在剩下的空格里面找，需要把剩下的行列补全

*/
/************************************************
* Author        :Powatr
* Created Time  :2015-8-17 21:05:24
* File Name     :B.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 50 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;

double dp[MAXN][MAXN][MAXN*MAXN];
int main(){
    int n, m, T;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &m);
        dp[0][0][0] = 1;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                for(int k = 1; k <= n*m; k++){
                    dp[i][j][k] = dp[i-1][j][k-1]*1.0*(j*(n-i+1))/(n*m-k+1)
                                + dp[i][j-1][k-1]*1.0*i*(m-j+1)/(n*m-k+1)
                                + dp[i-1][j-1][k-1]*1.0*(n-i+1)*(m-j+1)/(n*m-k+1)
                                + dp[i][j][k-1]*(1.0*i*j-k+1)/(n*m-k+1);
                }
            }
        }
        double ans = 0;
       for(int i = 1; i <= n*m; i++){
            ans += (dp[n][m][i] - dp[n][m][i-1])*i;
       }
       printf("%.10f\n", ans);
    }
    return 0;
}