zoj3822||牡丹江現場賽D題 機率dp

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

DominationTime Limit: 8 Seconds      Memory Limit: 131072 KB      Special Judge

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

21 32 2

Sample Output

3.0000000000002.666666666667

/**zoj 3822題意：問把一個n*m的棋盤填成每行至少一個棋子，每列至少一個棋子的狀態，需要棋子的數學期望解題思路： dp[i][j][k]表示到達結果狀態時所需要的棋子數的期望值。那麼dp[0][0][0]就是結果，狀態轉移方程見代碼。*/#include <stdio.h>#include <iostream>#include <string.h>#include <algorithm>using namespace std;int T,n,m;double dp[55][55][2555];int main(){    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        memset(dp,0,sizeof(dp));        for(int i=n; i>=0; i--)        {            for(int j=m; j>=0; j--)            {                if(j==m&&n==i)                    continue;                for(int t=i*j; t>=max(i,j); t--)                {                    dp[i][j][t]+=dp[i][j][t+1]*1.0*(i*j-t)/(n*m-t);                    dp[i][j][t]+=dp[i][j+1][t+1]*1.0*(m-j)*i/(n*m-t);                    dp[i][j][t]+=dp[i+1][j][t+1]*(1.0)*(n-i)*j/(n*m-t);                    dp[i][j][t]+=dp[i+1][j+1][t+1]*(1.0)*(n-i)*(m-j)/(n*m-t);                    dp[i][j][t]+=1.0;                }            }        }        printf("%.12lf\n",dp[0][0][0]);    }    return 0;}

zoj3822||牡丹江現場賽D題 機率dp