ZOJ3822 ACM-ICPC 2014 亚洲杯赛事现场牡丹江司D称号Domination 可能性DP

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

Sample Output

3.000000000000

2.666666666667

    这道题算是一个概率DP的裸题吧，非常裸的算法，题目意思是告诉你有一个n*m的区域，占据一个格子能够控制这个行和列，求控制全部行列所须要占据格子的期望，直接推导出状态转移方程就可以。对于算概率，我设dp[i][j][k]表示的意思为当我用k步控制了i行j列的概率，那么我们能够知道。这时候下一步则有4不状态转移。分别为继续取已控制的i行j列的格子，取未控制的i行，已控制的j列的格子，取已控制的i行和未控制的j列的格子，取未控制的i行j列的格子四种不同的情况，对于每种情况，状态转移方程例如以下：

1、继续取已控制的i行j列的格子

dp[i][j][k+1]+=dp[i][j][k]*(i*j-k)/(n*m-k);
2、取未控制的i行，已控制的j列的格子

dp[i+1][j][k+1]+=dp[i][j][k]*(n-i)*j/(n*m-k);
3、取已控制的i行和未控制的j列的格子

dp[i][j+1][k+1]+=dp[i][j][k]*(m-j)*i/(n*m-k);
4、取未控制的i行j列的格子

dp[i+1][j+1][k+1]+=dp[i][j][k]*(n-i)*(m-j)/(n*m-k);
从而终于求出控制n行m列须要步数的概率，由期望的定义式EX=n1p1+n2p2+...就可以求出终于期望，详细程序例如以下：

#include<cstdio>

#include<iostream>

#include<cstring>

#include<cmath>

#include<cstdlib>

#include<algorithm>

#include<map>

#include<vector>

#include<queue>

using namespace std;

double dp[55][55][55*55],a[55][55];

const double eps=1e-8;

int maxx(int a,int b)

{

    if(a>b)return a;

    return b;

}

int main()

{

  //  freopen("in.txt","r",stdin);

    int t;

    cin>>t;

    while(t--)

    {

        int n,m;

        scanf("%d%d",&n,&m);

        memset(dp,0,sizeof(dp));

        memset(a,0,sizeof(a));

        dp[1][1][1]=1;

        a[1][1]=1;

        for(int i=1;i<=n;i++)

        {

            for(int j=1;j<=m;j++)

            {

                for(int k=1;k<=a[i][j];k++)

                {

                    if(i==n&&j==m)

                        break;

                    dp[i][j][k+1]+=dp[i][j][k]*(i*j-k)/(n*m-k);

                    if(i*j>k)

                        a[i][j]=maxx(a[i][j],k+1);

                }

                for(int k=1;k<=a[i][j];k++)

                {

                    dp[i+1][j][k+1]+=dp[i][j][k]*(n-i)*j/(n*m-k);

                    a[i+1][j]=maxx(a[i+1][j],k+1);

                    dp[i][j+1][k+1]+=dp[i][j][k]*(m-j)*i/(n*m-k);

                    a[i][j+1]=maxx(a[i][j+1],k+1);

                    dp[i+1][j+1][k+1]+=dp[i][j][k]*(n-i)*(m-j)/(n*m-k);

                    a[i+1][j+1]=maxx(a[i+1][j+1],k+1);

                }

            }

        }

        double sum=0;

        for(int i=1;i<=a[n][m];i++)

            sum+=dp[n][m][i]*i;

        printf("%.12f\n",sum);

    }

    return 0;

}