zoj 3288 Domination (概率dp)

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///dp[i][j][k]表示i行j列已经有棋子，且放了k个的概率///dp[i][j][k]一共有四种转移方式///1：dp[i-1][j][k-1]  概率为 (n-(i-1))*j/(n*m-(k-1))///2:dp[i][j-1][k-1]   概率为  i*(m-(j-1))/(n*m-(k-1))///3:dp[i-1][j-1][k-1]  概率为 (n-(i-1))*(m-(j-1))/(n*m-(k-1))///4:dp[i][j][k-1]  概率为 (i*j-(k-1))/(n*m-(k-1))# include <stdio.h># include <algorithm># include <string.h># include <iostream>using namespace std;double  dp[55][55][2510];int main(){    int n,m,t,i,j,k;    double ans;    while(~scanf("%d",&t))    {        while(t--)        {            scanf("%d%d",&n,&m);            memset(dp,0,sizeof(dp));            dp[0][0][0]=1;            for(i=1; i<=n; i++)            {                for(j=1; j<=m; j++)                {                    for(k=1; k<=n*m; k++)                    {                        if(i==n&&j==m)                            dp[i][j][k]=dp[i-1][j][k-1]*(n-(i-1))*j/(n*m-(k-1))+dp[i][j-1][k-1]*i*(m-(j-1))/(n*m-(k-1))+dp[i-1][j-1][k-1]*(n-(i-1))*(m-(j-1))/(n*m-(k-1));                        else                            dp[i][j][k]=dp[i-1][j][k-1]*(n-(i-1))*j/(n*m-(k-1))+dp[i][j-1][k-1]*i*(m-(j-1))/(n*m-(k-1))+dp[i-1][j-1][k-1]*(n-(i-1))*(m-(j-1))/(n*m-(k-1))+dp[i][j][k-1]*(i*j-(k-1))/(n*m-(k-1));                    }                }            }            ans=0;            for(i=1; i<=n*m; i++) ///求期望==概率乘天数的和集                ans+=dp[n][m][i]*i;            printf("%.12lf\n",ans);        }    }    return 0;}
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