-----概率DP ZOJ 3822- Domination

最新推荐文章于 2019-08-09 14:35:50 发布

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本文介绍了一个关于棋盘覆盖的问题，即在N*M的棋盘上随机放置棋子，直到每行每列至少有一个棋子。文章通过概率DP的方法解决了这个问题，并给出了具体的实现代码。

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
题意：
给你一个N*M的棋盘，每天都会随机在棋盘上放一颗棋子，要求至少每行每列都要有一颗棋子，求满足条件下的天数的期望
解题思路：
概率DP
dp[i][j][k]表示，已经放了i颗棋子，覆盖了 j行k列的概率
然后推状态方程
这里写图片描述
由该图可知，在满足题目要求的条件之前,不论是哪个状态下都能分成如上四个区域，状态方程如下：
当前的概率=前一步的概率*当前的概率
区域一：放在旧的行和旧的列之中，只有棋子数增加，行数和列数保持不变
dp[i+1][j][k]+=dp[i][j][k]*(j*k-i)/(n*m-i)
区域二：新的行和旧的列，棋子数增加，行数增加，列数保持不变
dp[i+1][j+1][k]+=dp[i][j][k]*((n-j)*k)/(n*m-i)
区域三：旧的行和新的列，棋子数增加，行数保持不变，列数增加
dp[i+1][j][k+1]+=dp[i][j][k](j(m-k))/(n*m-i)
区域四：新的行和新的列，旗子数增加，行数增加，数数增加
dp[i+1][j+1][k+1]+=dp[i][j][k]((n-j)(m-k))/(n*m-i)
！！还要注意每个状态方程的条件

#include <iostream>
#include <cstring>
#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;

const int maxn=55;
double dp[maxn*maxn][maxn][maxn];
int n,m;

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        memset(dp,0,sizeof(dp));
        dp[1][1][1]=1;
        for(int i=1; i<n*m; i++)
            for(int j=1; j<=n; j++)
                for(int k=1; k<=m; k++)
                {
                    if(dp[i][j][k]==0||j==n&&k==m) continue;
                    if(j*k>i) dp[i+1][j][k]+=dp[i][j][k]*(j*k-i)/(n*m-i);
                    if(j<n) dp[i+1][j+1][k]+=dp[i][j][k]*((n-j)*k)/(n*m-i);
                    if(k<m) dp[i+1][j][k+1]+=dp[i][j][k]*(j*(m-k))/(n*m-i);
                    if(j<n&&k<m) dp[i+1][j+1][k+1]+=dp[i][j][k]*((n-j)*(m-k))/(n*m-i);
                }
        double ans=0;
        for(int i=1; i<=n*m; i++)
            ans+=i*dp[i][n][m];
            printf("%.12lf\n",ans);
    }
}