凸包性质——cf1044C

由题目给出的条件不难发现这是个凸包

又因为要求的是曼哈顿距离的最大值，所以当k>=4时都是同一个答案，即上下左右的最远点构成的凸包外接矩形

当k=3时固定两个极点，枚举第三个点即可(赋值粘贴就完事了)

#include<bits/stdc++.h>
using namespace std;
#define N 300005
#define ll long long  

struct point{
    ll x,y;
}p[N],l,r,u,d;
int n;
int cmp1(point &a,point &b){return a.x<b.x;}
int cmp2(point &a,point &b){return a.x>b.x;}
int cmp3(point &a,point &b){return a.y>b.y;}
int cmp4(point &a,point &b){return a.y<b.y;}
int equ(point a,point b){return a.x==b.x && a.y==b.y;}

int main(){
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>p[i].x>>p[i].y;
    sort(p+1,p+1+n,cmp1);l=p[1];
    sort(p+1,p+1+n,cmp2);r=p[1];
    sort(p+1,p+1+n,cmp3);u=p[1];
    sort(p+1,p+1+n,cmp4);d=p[1];
    
    ll Max=0;
    for(int i=1;i<=n;i++){
        if(equ(p[i],l)||equ(p[i],r))continue;
        point k1=p[i],k2=l,k3=r;
        ll up=max(k1.y,max(k2.y,k3.y));
        ll down=min(k1.y,min(k2.y,k3.y));
        ll ri=max(k1.x,max(k2.x,k3.x));
        ll le=min(k1.x,min(k2.x,k3.x));
        Max=max(Max,2*(ri-le)+2*(up-down));
    }
    for(int i=1;i<=n;i++){
        if(equ(p[i],u)||equ(p[i],d))continue;
        point k1=p[i],k2=u,k3=d;
        ll up=max(k1.y,max(k2.y,k3.y));
        ll down=min(k1.y,min(k2.y,k3.y));
        ll ri=max(k1.x,max(k2.x,k3.x));
        ll le=min(k1.x,min(k2.x,k3.x));
        Max=max(Max,2*(ri-le)+2*(up-down));    
    }
    for(int i=1;i<=n;i++){
        if(equ(p[i],l)||equ(p[i],d))continue;
        point k1=p[i],k2=l,k3=d;
        ll up=max(k1.y,max(k2.y,k3.y));
        ll down=min(k1.y,min(k2.y,k3.y));
        ll ri=max(k1.x,max(k2.x,k3.x));
        ll le=min(k1.x,min(k2.x,k3.x));
        Max=max(Max,2*(ri-le)+2*(up-down));
    }
    for(int i=1;i<=n;i++){
        if(equ(p[i],l)||equ(p[i],u))continue;
        point k1=p[i],k2=l,k3=u;
        ll up=max(k1.y,max(k2.y,k3.y));
        ll down=min(k1.y,min(k2.y,k3.y));
        ll ri=max(k1.x,max(k2.x,k3.x));
        ll le=min(k1.x,min(k2.x,k3.x));
        Max=max(Max,2*(ri-le)+2*(up-down));
    }
    for(int i=1;i<=n;i++){
        if(equ(p[i],r)||equ(p[i],d))continue;
        point k1=p[i],k2=r,k3=d;
        ll up=max(k1.y,max(k2.y,k3.y));
        ll down=min(k1.y,min(k2.y,k3.y));
        ll ri=max(k1.x,max(k2.x,k3.x));
        ll le=min(k1.x,min(k2.x,k3.x));
        Max=max(Max,2*(ri-le)+2*(up-down));
    }
    for(int i=1;i<=n;i++){
        if(equ(p[i],r)||equ(p[i],u))continue;
        point k1=p[i],k2=r,k3=u;
        ll up=max(k1.y,max(k2.y,k3.y));
        ll down=min(k1.y,min(k2.y,k3.y));
        ll ri=max(k1.x,max(k2.x,k3.x));
        ll le=min(k1.x,min(k2.x,k3.x));
        Max=max(Max,2*(ri-le)+2*(up-down));
    }

    cout<<Max<<" ";
    for(int i=4;i<=n;i++)
        cout<<2*(u.y-d.y)+2*(r.x-l.x)<<" ";
}