ZOJ 3822 Domination【概率dp】

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博客围绕一个棋盘放棋问题展开，每天在随机空单元格放一个棋子，要使棋盘每行每列至少有一个棋子。需计算放棋子天数的期望值，方法是求出每种天数的概率后套用公式，还给出了输入输出示例。

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3822

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

题意：

每天放一个棋子，使得每一行每一列至少一个棋子，问放棋子的天数的期望值

分析：

求出每种天数的概率，最后套用公式

代码：

#include<bits/stdc++.h>
using namespace std;
double dp[2501][51][51];///dp[k][i][j] 表示放k个棋子 占i行 和 j列的概率
int main()
{
    int t;
    int N, M;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &N, &M);
        memset(dp, 0, sizeof(dp));
        dp[0][0][0] = 1;
        double p;
        for(int k = 0; k <= N*M; k++)
        {
            for(int i = 0; i <= N; i++)
            {
                for(int j = 0; j <= M; j++)
                {
                    if(i == N && j == M || dp[k][i][j] == 0) continue;
                    if(i * j >= k)
                        dp[k+1][i][j] += dp[k][i][j] * (i * j - k) / (N * M - k);
                    if(i < N)
                        dp[k+1][i+1][j] += dp[k][i][j] * (N - i) * j / (N * M - k);
                    if(j < M)
                        dp[k+1][i][j+1] += dp[k][i][j] * (M - j) * i / (N * M - k);
                    if(i < N && j < M)
                        dp[k+1][i+1][j+1] += dp[k][i][j] * (N - i) * (M - j) / (N * M - k);
                }
            }
        }
        double ans = 0;
        for(int i = 0; i <= N*M; i++)//求期望，套公式概率*天数的和
            ans += i * dp[i][N][M];
        printf("%.10f\n", ans);
    }
    return 0;
}