zoj 3822 Domination（2014牡丹江区域赛D题）

最新推荐文章于 2020-10-04 21:07:00 发布

原创最新推荐文章于 2020-10-04 21:07:00 发布 · 1.9k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#algorithm

ACM-区域赛专栏收录该内容

22 篇文章

订阅专栏

爱德华在马尔加大学工作之余，购买了一个大小为N行M列的大装饰棋盘，并每天随机放置棋子。几天后，他发现棋盘被棋子完全占据。本文提供了一个算法帮助爱德华计算从空棋盘到完全占据所需的平均天数。

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

概率dp，i为第i天，j为有j行已经被覆盖，k为k列已经被覆盖，则每个dp[i][j][k]都可以由上dp[i-1]推出。

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
double dp[5000][55][55];
int main(){
    int m,n;
    int T;
    scanf("%d",&T);
    memset(dp, 0, sizeof(dp));
    dp[0][0][0]=1;

    while (T--) {
        scanf("%d %d",&m,&n);
        for(int i=1;i<=m*n+1-min(m, n);++i){
            for(int j=1;j<=i&&j<=m;++j){
                for(int k=1;k<=i&&k<=n;++k){
                    if(j>m){
                        dp[i][j][k]=0;
                        continue;
                    }
                    if(k>n) {
                        dp[i][j][k]=0;
                        continue;
                    }
                    if(j!=m||k!=n)
                    dp[i][j][k]=dp[i-1][j-1][k-1]*(double)((m+1-j)*(n+1-k))/(double)(m*n-i+1)+dp[i-1][j-1][k]*(double)(k*(m+1-j))/(double)(m*n-i+1)+dp[i-1][j][k-1]*(double)(j*(n+1-k))/(double)(m*n-i+1)+dp[i-1][j][k]*(double)(j*k+1-i)/(double)(m*n-i+1);
                    else
                    dp[i][j][k]=dp[i-1][j-1][k-1]*(double)((m+1-j)*(n+1-k))/(double)(m*n-i+1)+dp[i-1][j-1][k]*(double)(k*(m+1-j))/(double)(m*n-i+1)+dp[i-1][j][k-1]*(double)(j*(n+1-k))/(double)(m*n-i+1);
                }
            }
        }
        double result=0;
        for(int i=1;i<=m*n+1-min(m, n);++i){
            result+=i*dp[i][m][n];
        }
        printf("%.12f\n",result);
    }
    return 0;
}