LA 6972 Domination

6972 Domination 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：题意：
给一个n*m的矩阵，每天随机的在未放棋子的格子上放一个棋子。求每行至少有一个棋子，每列至少有一个棋子的天数的期望
 (1 <= N, M <= 50).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
#define SC  scanf
#define CT continue
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
const double pi=acos(-1);
const int mod=100000000;

double dp[55][55][2505];
int main()
{
    int cas,n,m;SC("%d",&cas);
    while(cas--){
        SC("%d%d",&n,&m);
        MM(dp,0);
        dp[0][0][0]=1;
        dp[1][1][1]=1;
        for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        for(int k=max(i,j);k<=i*j;k++){
            int t=n*m-(k-1);
            double p1=dp[i][j][k-1]*(i*j-(k-1))/t;
            double p2=dp[i-1][j][k-1]*(n-(i-1))*j/t;
            double p3=dp[i][j-1][k-1]*(m-(j-1))*i/t;
            double p4=dp[i-1][j-1][k-1]*(n-(i-1))*(m-(j-1))/t;
            dp[i][j][k]=p1+p2+p3+p4;
        }
        double ans=0;
        for(int i=max(n,m);i<=n*m;i++)
            ans+=i*(dp[n][m][i]-dp[n][m][i-1]);
        printf("%.12f\n",ans);
    }
    return 0;
}

　　题解：