zoj 3822 Domination

zoj 3822 Domination

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

題解及代碼：

比賽的時候，這道題目沒寫出來。用dp[i][j][k]代表狀態為使用k個棋子覆蓋i行j列的概率，一開始的時候公式推導錯了，實際上應該是下面這樣的：

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);

同時比賽的時候也沒能想到(i==m&&j==n)的情況要減去藍色的部分，以為當滿足m行n列之後還能繼續放棋子。

#include 
#include 
#include 
#include 
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;
}