ZOJ3822-Domination 概率DP

Domination

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Time Limit: 8 Seconds      Memory Limit: 131072 KB      Special Judge

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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 withN rows andM columns.

Every day after work, Edward will place a chess piece on a random empty cell. A few days later, he found the chessboard wasdominated 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 ofN ×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

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Author: JIANG, Kai
Source: The 2014 ACM-ICPC Asia Mudanjiang Regional Contest

这道题是一个很裸的概率DP。题目意思是告诉你有一个n*m的区域，占领一个格子可以控制这个行和列，求控制所有行列所需要占领格子的期望。dp[i][j][k]表示有i行、j列在k步时有了棋子的概率，那么这个的概率就是它第k步走了ij都已被占据过的格子的概率+i被占据、j未被占据的概率+i未被占据、j被占据的概率+i、j均未被占据的概率。

注意：当i、j同时到n、m后，就结束了

#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
using namespace std;

double dp[55][55][2505]; //有i行、j列在k步时有了棋子的概率 
double max(double x,double y){
	if(x-y > 1e-6) return x;
	return y;
}
int main(){
	int T;
	scanf("%d", &T);
	while(T --){
		int n,m;
		scanf("%d%d", &n, &m);
		int t = n*m;
		memset(dp,0,sizeof(dp));
		dp[1][1][1] = 1.0;
		for(int i = 1;i <= n;i++){
			for(int j = 1;j <= m;j++){
				for(int k = 1;k <= t;k++){
					if(t-k+1 > 0){
						if(!(i==n && j==m) && i*j-k+1>0)	dp[i][j][k] += dp[i][j][k-1]*(i*j-k+1)/(t-k+1); //注意如果已经全部占据了i、j就已经结束了 
						if(n-i+1 > 0)	dp[i][j][k] += dp[i-1][j][k-1]*((n-i+1)*j)/(t-k+1);
						if(m-j+1 > 0)	dp[i][j][k] += dp[i][j-1][k-1]*(i*(m-j+1))/(t-k+1);
						if(n-i+1>0 && m-j+1>0)	dp[i][j][k] += dp[i-1][j-1][k-1]*((n-i+1)*(m-j+1))/(t-k+1);
					}
				}
			}
		}
		double ans = 0;
		for(int k = 1;k <= t;k++){
			ans += dp[n][m][k]*k;
		}
		printf("%.12lf\n", ans);
	}
	return 0;
}