「CF398B」Painting The Wall

2015年2月1日4,0390

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.

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.
User ainta choose any tile on the wall with uniform probability.
If the tile he has chosen is not painted, he paints the tile. Otherwise, he ignores it.
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 r_i and c_i (1 ≤ r_i, c_i ≤ 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 from 1 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

5 2

2 3

4 1

output

11.7669491886

1	11.7669491886

input

2 2
1 1
1 2

2 2

1 1

1 2

output

2.0000000000

1	2.0000000000

input

1 1
1 1

1 1

output

0.0000000000

1	0.0000000000

题解

dp[x][y]表示还剩x行y列需要染色，期望达到目标状态的步数

dp[x][y]=(x*y*dp[x-1][y-1]+x*(n-y)*dp[x-1][y]+y*(n-x)*dp[x][y-1]+dp[x][y]*(n-x)(n-y))/(n*n)

移项即可得到方程

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<set>
#include<ctime>
#include<vector>
#include<queue>
#include<algorithm>
#include<map>
#include<cmath>
#define inf 1000000000
#define rad 100000000
#define pa pair<int,int>
#define ll long long 
using namespace std;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
double dp[2005][2005];
int n,m,t1,t2;
bool r[2005],c[2005];
int main()
{
	n=read();m=read();
	t1=t2=n;
	for(int i=1;i<=m;i++)
	{
		int x=read(),y=read();
		if(!r[x])r[x]=1,t1--;
		if(!c[y])c[y]=1,t2--;
	}
	dp[0][0]=0;
	for(int i=1;i<=n;i++)
	{
		dp[i][0]=dp[i-1][0]+(double)n/i;
		dp[0][i]=dp[0][i-1]+(double)n/i;
	}
	for(int i=1;i<=t1;i++)
		for(int j=1;j<=t2;j++)
	    {
			dp[i][j]=n*n;
			dp[i][j]+=i*j*dp[i-1][j-1];
			dp[i][j]+=i*(n-j)*dp[i-1][j];
			dp[i][j]+=j*(n-i)*dp[i][j-1];
			dp[i][j]/=(n*n-(n-i)*(n-j));
		}
	printf("%.10lf\n",dp[t1][t2]);
	return 0;
}

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<set>

#include<ctime>

#include<vector>

#include<queue>

#include<algorithm>

#include<map>

#include<cmath>

#define inf 1000000000

#define rad 100000000

#define pa pair<int,int>

#define ll long long

using namespace std;

int read()

{

int x=0,f=1;char ch=getchar();

while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

return x*f;

}

double dp[2005][2005];

int n,m,t1,t2;

bool r[2005],c[2005];

int main()

{

n=read();m=read();

t1=t2=n;

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

{

int x=read(),y=read();

if(!r[x])r[x]=1,t1--;

if(!c[y])c[y]=1,t2--;

}

dp[0][0]=0;

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

{

dp[i][0]=dp[i-1][0]+(double)n/i;

dp[0][i]=dp[0][i-1]+(double)n/i;

}

for(int i=1;i<=t1;i++)

for(int j=1;j<=t2;j++)

{

dp[i][j]=n*n;

dp[i][j]+=i*j*dp[i-1][j-1];

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

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

dp[i][j]/=(n*n-(n-i)*(n-j));

}

printf("%.10lf\n",dp[t1][t2]);

return 0;

}

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「CF398B」Painting The Wall

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