ZOJ3822 ACM-ICPC 2014 亚洲区域赛牡丹江赛区现场赛D题Domination 概率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

2
1 3
2 2

Sample Output

3.000000000000
2.666666666667

这道题算是一个概率DP的裸题吧，很裸的算法，题目意思是告诉你有一个n*m的区域，占领一个格子可以控制这个行和列，求控制所有行列所需要占领格子的期望，直接推导出状态转移方程即可，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，那么我们可以知道，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，状态转移方程如下：

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);

从而最终求出控制n行m列需要步数的概率，由期望的定义式EX=n1p1+n2p2+...即可求出最终期望，具体程序如下：

#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
[m];i++)
sum+=dp
[m][i]*i;
printf("%.12f\n",sum);
}
return 0;
}