zoj 3822 概率期望dp入门

最新推荐文章于 2019-05-06 22:42:24 发布

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最新推荐文章于 2019-05-06 22:42:24 发布

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原文链接：http://www.cnblogs.com/lastone/p/5412518.html

本文探讨了一个关于棋盘占用概率的问题，通过数学建模和动态规划算法，计算了达到特定状态所需的平均天数。

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Description

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 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 of N × Mdominated. 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：有一个n*m的棋盘，一个无聊的人每天早上在棋盘空白处随机放一颗棋子，有一天他发现所有行和列都至少有一个棋子（他每天都检查一遍棋盘是否满足这个条件），求天数的期望。

dp[i][j][k]：到第i天，棋子占据j行，k列的概率。

状态转移方程比较好想，但是这里需要加一个条件，那就是计算dp[i][n][m]这个状态时，之前i-1天都不能使棋盘占据n行m列。

如果不加上这个条件，可能会在之前i-1天就占据了n行m列，其实这正是我之前的写法，这并不是最早发现占据n行m列的天数，所以不能直接用来求期望，一开始我的解决方法是在算完dp之后又来计算ans[i]，它表示到第i天第一次占据n行m列的概率，但这样算出来是错误的（明显是错误的）。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
double dp[3000][55][55];
int main() {
    int T,n,m;
    scanf("%d",&T);
    while(T--) {
        memset(dp,0,sizeof(dp));
        dp[0][0][0] = 1;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n*m;i++) {
            for(int j=1;j<=n;j++)
            for(int k=1;k<=m;k++) {
                if(j!=n||k!=m)
                dp[i][j][k] += dp[i-1][j][k]*(j*k-(i-1))/(m*n-(i-1));
                dp[i][j][k] += dp[i-1][j-1][k-1]*(m*n-m*(j-1)-n*(k-1)+(j-1)*(k-1))/(m*n-(i-1));
                dp[i][j][k] += dp[i-1][j][k-1]*(j*m-j*(k-1))/(m*n-(i-1));
                dp[i][j][k] += dp[i-1][j-1][k]*(k*n-k*(j-1))/(m*n-(i-1));
            }
        }
        double r=0;
        for(int i=1;i<=n*m;i++) {
            r += dp[i][n][m]*i;
        }
        printf("%.12f\n",r);
    }
}