ZOJ 3826 Hierarchical Notation(2014 牡丹江 D，概率DP)

DominationTime Limit: 8 Seconds Memory Limit: 131072 KB Special JudgeEdward 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 withN rows andM columns.Every day after work, Edward will place a chess piece on a random empty cell. A few days later, he found the chessboard wasdominated 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 ofN ×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

题目大意：有一个N*M的棋盘，每次随机的在一个没有落棋子的点落一个棋子。然后求每一行每一列都最少有一颗棋子的期望步数。解题思路：定义dp[k][i][j]表示已经走了k步，并且i行j列的格子都有了棋子。那么剩余可以落子的位置有n*m-k个。那么一共会有4种情况：（1）落在一个行列都没有棋子的位置 那么dp[k+1][i+1][j+1]+=dp[k][i][j]*(m-j)*(n-i)/(n*m-k);//就是那n-i行m-j列的位置可以落子（2）落在一个行有棋子，列没棋子的位置， 那么dp[k+1][i][j+1]+=dp[k][i][j]*((m-j)*i)/(n*m-k);//就是m-j列 n行。（3）落在一个列有棋子，行没棋子的位置， 那么dp[k+1][i+1][j]+=dp[k][i][j]*((n-i)*j)/(m*n-k);//同上（4）落在行列都有棋子的位置 那么这种转移的条件就是i*j>k 并且i != n并且 j != m(因为那样是不能落子的)所以dp[k+1][i][j]+=dp[k][i][j]*(i*j-k)/(m*n-k);

<pre name="code" class="cpp">#include<cstdio>#include<cmath>#include<cstring>#define eps 1e-12double dp[2510][55][55];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m;        scanf("%d %d",&n,&m);        memset(dp,0,sizeof(dp));        dp[0][0][0]=1;        double ans=0;        for(int k=0;k<=n*m;k++)        {            for(int i=0;i<=n;i++)                for(int j=0;j<=m;j++)                {                    if(i*j>k&&(i!=n||j!=m))                        dp[k+1][i][j]+=dp[k][i][j]*(i*j-k)/(double)(m*n-k);                    dp[k+1][i+1][j+1]+=dp[k][i][j]*(m-j)*(n-i)/(double)(n*m-k);                    dp[k+1][i+1][j]+=dp[k][i][j]*((n-i)*j)/(double)(m*n-k);                    dp[k+1][i][j+1]+=dp[k][i][j]*((m-j)*i)/(double)(n*m-k);                }            ans+=dp[k][m]*k;        }        printf("%.12lf\n",ans);    }    return 0;}