zoj 3822概率dp

Domination

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

分析： 这是2014年牡丹江现场赛一题==

可以顺着推概率，最后E=p1*n1+p2*n2…..

也可以逆着推期望。

我用的是dp[i][j][k]表示有i行，j列在第k天被控制好了。

那么当第k+1天放棋子的时候就有四种状态转移；

dp[i][j][k+1]*p1

dp[i+1][j][k+1]*p2

dp[i][j+1][k+1]*p3

dp[i+1][j+1][k+1]*p4

这里的概率分别求就行~

#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <algorithm>
const int N=66*66;
using namespace std;
int n,m;
double dp[55][55][N];
//dp[i][j][k]表示有i行满足每行至少有一个棋子的条件，有j列满足每列至少有一个棋子的条件，共占据了K个格子

int main()
{
   int T;
   scanf("%d",&T);
   while(T--)
   {
     scanf("%d%d",&n,&m);
     int total=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=max(i,j);k<=i*j;k++)
        {
             if(i==n && j==m)break;
             //cout<<i<<" "<<j<<" "<<k<<"  ";
             //printf("%.12lf\n",dp[i][j][k]);
             if(k!=i*j)dp[i][j][k+1]+=(i*j-k)*1.0/(total-k)*dp[i][j][k];
             if(j+1<=m)
             {
              //cout<<i<<" "<<j<<" "<<k<<"  ";
              dp[i][j+1][k+1]+=(i*m-i*j)*1.0/(total-k)*dp[i][j][k];
              // printf("%.12lf\n",dp[i][j+1][k+1]);

             }

             if(i+1<=n)
             {
                dp[i+1][j][k+1]+=(j*n-i*j)*1.0/(total-k)*dp[i][j][k];
             }
             if(i+1<=n && j+1<=m)
             {
              dp[i+1][j+1][k+1]+=(total-i*m-j*n+i*j)*1.0/(total-k)*dp[i][j][k];
             }
        }

        int cnt=max(n*(m-1),m*(n-1))+1;
        double ans=0.0;
        for(int i=max(n,m);i<=cnt;i++)
        {
               ans+=dp[n][m][i]*i;
               //cout<<n<<" "<<m<<" "<<i<<"  ";
               //printf("%.12lf\n",dp[n][m][i]);
               //printf("%.12lf\n",ans);
        }
        printf("%.12lf\n",ans);
   }
   return 0;
}