中位数 贪心 

x,y 分开考虑

显然中间那个点一定最省 

1.

奇数 显然中间那个点一定最省  之后走一步答案必会增加，因此中间和4个方向一定最省，另外4个方向增加是对应2个方向的和

2

偶数

判定整个范围会超时

枚举牛而非整个区间，否则TLE

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<functional>
#include<algorithm>
#define MAXN (10000 + 10)
#define INF 400000000 + 10
using namespace std;
int x[MAXN],y[MAXN];
int a[4][2]={{1,0},{-1,0},{0,-1},{0,1}};
int node[MAXN][2];
int n;
int sum(int nx,int ny)
{
	int tot=0;
	for (int i=1;i<=n;i++)
	{
		tot+=abs(x[i]-nx)+abs(y[i]-ny);
	}
	return tot;
}
int b(int x,int y)
{
	for (int i=1;i<=n;i++)
		if (node[i][0]==x&node[i][1]==y) return true;
	return false;
}
void find(int &minway,int &ans,int x,int y)
{
	int nowtot=sum(x,y);
	if (nowtot<minway)
	{
		ans=1;
		minway=nowtot;
	}
	else if (nowtot==minway) ans++;

}


int calc()
{
	int m=(1+n)/2;
	if (n%2)
	{
		if (!b(x[m],y[m]))
		{
			printf("%d",sum(x[m],y[m]));
			return 1;
		}
		else
		{
			int minway=INF,ans=0;
			for (int i=0;i<4;i++)
			{
				find(minway,ans,x[m]+a[i][0],y[m]+a[i][1]);
			}
			printf("%d",minway);
			return ans;
		}

	}
	else
	{
		int ans=0;
		for (int i=1;i<=n;i++)
		{
			if (x[m]<=node[i][0]&&node[i][0]<=x[m+1]&&y[m]<=node[i][1]&&node[i][1]<=y[m+1]) ans++;
		}
		ans=(x[m+1]-x[m]+1)*(y[m+1]-y[m]+1)-ans;
		printf("%d",sum(x[m],y[m]));
		return ans;
	}
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%d%d",&x[i],&y[i]);
		node[i][0]=x[i];
		node[i][1]=y[i];
	}
	sort(x+1,x+n+1);
	sort(y+1,y+n+1);
	printf(" %dn",calc());
	return 0;
}