zoj 3822 dp

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5376

很变态的题，单本人觉得此题写的太好了

一下是我的代码

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


 const int sizen=55;
 double dp[sizen*sizen][sizen][sizen];
 int T;


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