ZOJ 3822 Domination(概率DP)

题目链接

Domination
Time 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

2
1 3
2 2
Sample Output

3.000000000000
2.666666666667

题目大意：
有一个n*m的棋盘，一个人每次往棋盘里面放一个棋子，然后当每一行和每一列中都有棋子的时候就不再放了，求放的次数的期望.

解题思路：
一看是一个概率题首先就得想概率DP啊，然后就是DP的表示了 ，怎么表示呢，只要想出DP的表示形式基本上这个题就已经成功了一半了，设DP[i][j][k]表示的是 有k个棋子覆盖了i行j列的概率（注意是概率），那么它的前一个状态是什么呢：
有四个状态，它的前一个状态肯定是由k-1个石子组成的，那么已经有k-1个石子了，只要再放一个石子，然而放的这个石子它可能造成四种结果，
1)增加一行 –>DP[i-1][j][k-1]
2)增加一列 –>DP[i][j-1][k-1]
3)增加一行同时增加一列 –>DP[i-1][j-1][k-1]
4)不增加行数也不增加列数 –>DP[i][j][k-1]
那么我们只要求出每一个状态的概率然后相加就行了，令sum=m*n
1)的概率就是：(n-i+1)*j/(sum-k+1)
2)的概率就是：(m-j+1)*i/(sum-k+1)
3)的概率就是：(n-i+1)*(m-j+1)/(sum-k+1)
4)的概率就是：(i*j-k+1)/(sum-k+1)
然后DP[i][j][k]就等于各自的状态乘以各自的概率，注意当i==n&&j==m 的时候，就不计算它的前一个状态了，因为题目要求的就是这样的。最后我们在求期望的时候，就是求DP[n][m][k] (1<=k<=n*m)的累加每次累加乘以k，概率乘以次数就是期望：

My Code:

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
double dp[55][55][55*55];
int main()
{
    int T, m, n;
    double p1, p2, p3, p4;
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        int sum = m*n;
        memset(dp, 0, sizeof(dp));
        dp[0][0][0] = 1;
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {
                for(int k=1; k<=sum; k++)
                {
                    if(i==n && j==m)///n行m列都有棋子，题目描述不用计算了
                    {
                        p1 = 1.0*(n-i+1)*j/(sum-k+1);///增加行的概率
                        p2 = 1.0*(m-j+1)*i/(sum-k+1);///增加列的概率
                        p3 = 1.0*(n-i+1)*(m-j+1)/(sum-k+1);///增加行和列的概率
                        dp[i][j][k] = dp[i-1][j][k-1]*p1+dp[i][j-1][k-1]*p2+dp[i-1][j-1][k-1]*p3;
                    }
                    else
                    {
                        p1 = 1.0*(n-i+1)*j/(sum-k+1);///增加行的概率
                        p2 = 1.0*(m-j+1)*i/(sum-k+1);///增加列的概率
                        p3 = 1.0*(n-i+1)*(m-j+1)/(sum-k+1);///增加行和列的概率
                        p4 = 1.0*(i*j-k+1)/(sum-k+1);///不增加行和列的概率
                        dp[i][j][k] = dp[i-1][j][k-1]*p1+dp[i][j-1][k-1]*p2+dp[i-1][j-1][k-1]*p3+dp[i][j][k-1]*p4;
                    }
                }
            }
        }
        double ret = 0;
        for(int i=1; i<=sum; i++)
            ret += dp[n][m][i]*i;
        printf("%.10lf\n",ret);
    }
    return 0;
}