概率dp，一开始用了二维超时，后来加了一位记忆化就不超时了啊。dp[x][y][z]代表已经覆盖了第x行y列此时还剩下k个空格。

所以：dp[x][y][z] = p1*dp[x+1][y][z-1]+p2*dp[x][y+1][z-1]+p3*dp[x+1][y+1][z-1]+p4*dp[x][y][z-1] + 1。

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 Mcolumns.

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

#include <set>
#include <map>
#include <queue>
#include <math.h>
#include <vector>
#include <string>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

#define eps 1e-9
#define pi acos(-1.0)
#define inf 107374182
#define inf64 1152921504606846976
#define lc l,m,tr<<1
#define rc m + 1,r,tr<<1|1
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (SIZE))
#define clearall(A, X) memset(A, X, sizeof(A))
#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))
#define memcopyall(A, X) memcpy(A , X ,sizeof(X))
#define max( x, y )  ( ((x) > (y)) ? (x) : (y) )
#define min( x, y )  ( ((x) < (y)) ? (x) : (y) )

using namespace std;

const int maxn = 55;

double dp[maxn][maxn][maxn*maxn];
int n, m;
int N;
double dfs(int x, int y, int z)
{
    if(dp[x][y][z] != -1.0) return dp[x][y][z];
    if(x == n && y == m)
    {
        dp[x][y][z] = 0.0;
        return dp[x][y][z];
    }
    double p1 = 0.0, p2 = 0.0, p3 = 0.0, p4 = 0.0;
    dp[x][y][z] = 1.0;
    if(x+1 <= n)
    {
        p1 = ((n-x)*y*1.0)/z*1.0;
        dp[x][y][z] += p1*dfs(x+1, y, z-1);
    }
    if(y+1 <= m)
    {
        p2 = ((m-y)*x*1.0)/z*1.0;
        dp[x][y][z] += p2*dfs(x, y+1, z-1);
    }
    if(x+1 <= n && y+1 <= m)
    {
        p3 = ((n-x)*(m-y))*1.0/z*1.0;
        dp[x][y][z] += p3*dfs(x+1, y+1, z-1);
    }
    p4 = 1.0-p1-p2-p3;
    if(p4 >= 0) dp[x][y][z] += p4*dfs(x, y, z-1);
    return dp[x][y][z];
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n, &m);
        N = n*m;
        if(n == 1 || m == 1)
        {
            printf("%.10lf\n",(double)max(n, m));
            continue;
        }
        for(int i = 0; i <= n; i++)
            for(int j = 0; j <= m; j++)
                for(int k = 0; k <= n*m; k++) dp[i][j][k] = -1.0;
        printf("%.10lf\n",dfs(1, 1, N-1)+1);
    }
    return 0;
}

ZOJ 3822 Domination(概率dp)