对于这题，我们可以得到d(i,j)=max(|xi-xj|,|yi-yj|)，
设x'=(x+y)/2,y'=(x-y)/2
那么d(i,j)=|xi'-xj'|+|yi'-yj'|
x，y轴可以分开统计
先是统计x轴，将所有松鼠的x'排序
用前缀和和后缀和求出某个松鼠到其他松鼠的X轴距离
Xi*(i-1)-sum(1..i-1)+sum(i+1..n)-(n-i)*Xi
y轴同理
#include<iostream>
#include<cstdio>
#include<algorithm>
#define N 100005
#define ll long long
using namespace std;
int n;
struct point{ll x,y;int num;}a[N];
double ans=1e20;
ll sx,sy,ax[N],bx[N],ay[N],by[N];
bool cmpx(point a,point b){return a.x<b.x;}
bool cmpy(point a,point b){return a.y<b.y;}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
ll x,y;
scanf("%lld%lld",&x,&y);
a[i].x=x+y;a[i].y=x-y;
}
sort(a+1,a+n+1,cmpy);
for(int i=1;i<=n;i++)
{
ay[i]=ay[i-1]+a[i].y;
by[n-i+1]=by[n-i+2]-a[n-i+1].y;
a[i].num=i;
}
sort(a+1,a+n+1,cmpx);
for(int i=1;i<=n;i++)
{
ax[i]=ax[i-1]+a[i].x;
bx[n-i+1]=bx[n-i+2]-a[n-i+1].x;
}
for(int i=1;i<=n;i++)
{
ll tmp=0;
tmp+=ax[n]-ax[i]-(n-i)*(ax[i]-ax[i-1]);
tmp+=bx[1]-bx[i]-(i-1)*(bx[i]-bx[i+1]);
int j=a[i].num;
tmp+=ay[n]-ay[j]-(n-j)*(ay[j]-ay[j-1]);
tmp+=by[1]-by[j]-(j-1)*(by[j]-by[j+1]);
if(tmp<ans)ans=tmp;
}
printf("%.0lf",ans/2);
return 0;
}