Codeforces Round #233 (Div. 2)D. Painting The Wall 概率DP

                                                                               D. Painting The Wall

 

User ainta decided to paint a wall. The wall consists of n2 tiles, that are arranged in an n × n table. Some tiles are painted, and the others are not. As he wants to paint it beautifully, he will follow the rules below.

1. Firstly user ainta looks at the wall. If there is at least one painted cell on each row and at least one painted cell on each column, he stops coloring. Otherwise, he goes to step 2.
2. User ainta choose any tile on the wall with uniform probability.
3. If the tile he has chosen is not painted, he paints the tile. Otherwise, he ignores it.
4. Then he takes a rest for one minute even if he doesn't paint the tile. And then ainta goes to step 1.

 

However ainta is worried if it would take too much time to finish this work. So he wants to calculate the expected time needed to paint the wall by the method above. Help him find the expected time. You can assume that choosing and painting any tile consumes no time at all.

Input

The first line contains two integers n and m (1 ≤ n ≤ 2·103; 0 ≤ m ≤ min(n2, 2·104)) — the size of the wall and the number of painted cells.

Next m lines goes, each contains two integers ri and ci (1 ≤ ri, ci ≤ n) — the position of the painted cell. It is guaranteed that the positions are all distinct. Consider the rows of the table are numbered from 1 to n. Consider the columns of the table are numbered from1 to n.

Output

In a single line print the expected time to paint the wall in minutes. Your answer will be considered correct if it has at most 10 - 4 absolute or relative error.

Sample test(s)

input

5 2
2 3
4 1

output

11.7669491886

input

2 2
1 1
1 2

output

2.0000000000

input

1 1
1 1

output

0.0000000000

题意：有一个n*n的墙，现在小明来刷墙，如果每一行每一列都至少有一个格子刷过了就停止工作，否则每次随机选一个格子，如果刷过了就不刷如果没刷过就刷，然后休息一分钟，求停止工作时时间的数学期望（开始之前已经有m个格子刷过了）
题解：dp[i][j]表示还有i行j列未刷
     初始化： dp[i][0]=((n-i)/n)*dp[i][0]+dp[i-1][0]*i/n+1;
dp[0][j]=((n-j)/n)*dp[0][j]+dp[0][j-1]*j/n+1;
     转移：   dp[i][j]=dp[i][j]*(n-i)(n-j)/n^2+dp[i-1][j]*(i*(n-j))/n^2+dp[i][j-1]*((n-i)*j)/n^2+dp[i-1][j-1]*(i*j)/n^2+1;

#include
#include
using namespace std;
double dp[2010][2010];
int n,m,a[2010],b[2010];
int main()
{
    cin>>n>>m;
    int x,y;
    int l=n,r=n;
    for(int i=0; i)
    {
        cin>>x>>y;
        if(!a[x]) l--;
        if(!b[y]) r--;
        a[x]=1,b[y]=1;
    }
    for(int i=1; i<=n; i++) dp[i][0]=dp[i-1][0]+(double)n/i;
    for(int j=1; j<=n; j++) dp[0][j]=dp[0][j-1]+(double)n/j;
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {dp[i][j]=(dp[i-1][j]*i*(n-j)+n*n+dp[i][j-1]*j*(n-i)+dp[i-1][j-1]*i*j)/(n*n-(n-i)*(n-j));
        }
    }
    printf("%0.10f\n",dp[l][r]);
    return 0;
}

代码

 

转载于:https://www.cnblogs.com/zxhl/p/4846283.html