题目链接： problemId=5376">http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5376
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
PS:
附上bin神的概率dp总结 Orz；
http://www.cnblogs.com/kuangbin/archive/2012/10/02/2710606.html
代码例如以下：（学习：http://blog.****.net/napoleon_acm/article/details/40020297）
//dp[i][j][k] 表示当前用了<=k个chess ,覆盖了i行j列(i*j的格子 每行至少一个。每列至少一个)的概率。
//
//dp[i][j][k] 由 dp[i][j][k-1] , dp[i-1][j][k-1], dp[i][j-1][k-1], dp[i-1][j-1][k-1]得到，
//分别表示 1、加入的新的一个chess, 2、不覆盖新的行列， 3、仅仅新覆盖一行。 仅仅新覆盖一列。
//4、同一时候新覆盖一行和一列,得到dp[i][j][k]。
//递推时。 每一个概率 * (能够覆盖的点数/剩余全部的空点数) 相加得到[i][j][k].
//ans += (dp[n][m][i] - dp[n][m][i-1])* i。 (i = [1, n*m])
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 57;
double dp[maxn][maxn][maxn*maxn];
int main()
{
int n, m;
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&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 <= n*m; k++)
{
dp[i][j][k] = dp[i][j-1][k-1]*((1.0*i*(m-j+1))/(n*m-k+1))
+dp[i-1][j][k-1]*((1.0*(n-i+1)*j)/(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("%.12lf\n",ans);
}
return 0;
}