zoj 3822 Domination

最新推荐文章于 2020-04-16 18:54:23 发布

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最新推荐文章于 2020-04-16 18:54:23 发布

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分类专栏：动态规划

本文链接：https://blog.youkuaiyun.com/knight_kaka/article/details/40052057

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本文介绍了一个关于放置棋子使棋盘每一行每一列至少有一个棋子的问题，并提供了一种利用动态规划解决该问题的方法。文章详细解释了状态转移方程，并给出了完整的代码实现。

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 Nrows 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

题解及代码：

比赛的时候，这道题目没写出来。用dp[i][j][k]代表状态为使用k个棋子覆盖i行j列的概率，一开始的时候公式推导错了，实际上应该是下面这样的：

i==m&&j==n dp[i][j][k]=(m-i+1)*(n-j+1)*dp[i-1][j-1][k-1]+j*(m-i+1)*1.0*dp[i-1][j][k-1]+i*(n-j+1)*1.0*dp[i][j-1][k-1];dp[i][j][k]/=(m*n-k+1);

else dp[i][j][k]=(m-i+1)*(n-j+1)*dp[i-1][j-1][k-1]+j*(m-i+1)*1.0*dp[i-1][j][k-1]+i*(n-j+1)*1.0*dp[i][j-1][k-1]+(i*j-k+1)*1.0*dp[i][j][k-1];dp[i][j][k]/=(m*n-k+1);

同时比赛的时候也没能想到(i==m&&j==n)的情况要减去蓝色的部分，以为当满足m行n列之后还能继续放棋子。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

double dp[55][55][2555];

int main()
{
    double base;
    int cas,n,m;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d%d",&m,&n);
        int maxn=n*m;
        memset(dp,0,sizeof(dp));
        dp[0][0][0]=1;
        for(int i=1;i<=m;i++)
        for(int j=1;j<=n;j++)
        {
            for(int k=1;k<=maxn;k++)
            {
                base=(double)(maxn-k+1);
                dp[i][j][k]+=(double)(m-i+1)*(n-j+1)*1.0*dp[i-1][j-1][k-1];
                dp[i][j][k]+=(double)j*(m-i+1)*1.0*dp[i-1][j][k-1];
                dp[i][j][k]+=(double)i*(n-j+1)*1.0*dp[i][j-1][k-1];
                if(i!=m||j!=n)
                dp[i][j][k]+=(double)(i*j-k+1)*1.0*dp[i][j][k-1];
                dp[i][j][k]/=base;
                //printf("dp[%d][%d][%d]:%.6lf\n",i,j,k,dp[i][j][k]);
            }
        }

        double ans=0;
        for(int i=1;i<=maxn;i++)
        {
            ans+=dp[m][n][i]*i;
        }
        printf("%.10lf\n",ans);
    }
    return 0;
}