Domination（概率DP)

最新推荐文章于 2018-07-15 17:34:00 发布

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本文介绍了一个关于放置棋子使棋盘每行每列至少有一个棋子的问题，并提供了一段C++代码来计算期望天数。该问题源于ACM竞赛，通过动态规划求解。

Link：http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5354

Domination

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

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
Source: The 2014 ACM-ICPC Asia Mudanjiang Regional Contest

AC code：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
using namespace std;
int n,m;
double dp[55][55][55*55];
void solve()
{
    memset(dp,0,sizeof(dp));
    dp[0][0][0]=1.0;
    int i,j,k;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
        {
            for(k=1;k<=n*m;k++)
            {
                dp[i][j][k]=dp[i-1][j-1][k-1]*((n-i+1)*(m-j+1))/(n*m-k+1);
                dp[i][j][k]+=dp[i-1][j][k-1]*(n-i+1)*j/(n*m-k+1);
                dp[i][j][k]+=dp[i][j-1][k-1]*(m-j+1)*i/(n*m-k+1);
                if(i!=n||j!=m)
                    dp[i][j][k]+=dp[i][j][k-1]*(i*j-k+1)/(n*m-k+1);
            }
        }
    }
    double ans=0;
    for(k=1;k<=n*m;k++)
    {
        ans+=(dp[n][m][k]*k);
    }
    printf("%.12lf\n",ans);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        solve();
    }
    return 0;
}