题目大意：给n个点，求点对最大距离的平方。

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很容易证明最大距离的点对在最大凸包上。

那么就是旋转卡壳的裸题了。

旋转卡壳要不要写讲解呢……

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            using namespace std;const int N=50001;struct point{ int x; int y;}p[N],q[N];int n,per[N],l;inline point getmag(point a,point b){ point s; s.x=b.x-a.x;s.y=b.y-a.y; return s;}inline int multiX(point a,point b){ return a.x*b.y-b.x*a.y;}inline int dis(point a,point b){ return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}inline bool cmp(int u,int v){ int det=multiX(getmag(p[1],p[u]),getmag(p[1],p[v])); if(det!=0)return det>0; return dis(p[1],p[u])
            
             =2&&multiX(getmag(q[l-1],p[j]),getmag(q[l-1],q[l]))>=0)l--; q[++l]=p[j]; } return;}inline int area(point a,point b,point c){ return multiX(getmag(a,b),getmag(a,c));}int calliper(){ if(l==2)return dis(q[1],q[2]); int res=0; for(int i=1,j=3;i<=l;i++){ while(j%l+1!=i&&area(q[i],q[i%l+1],q[j])<=area(q[i],q[i%l+1],q[j%l+1]))j=j%l+1; res=max(res,dis(q[i],q[j])); res=max(res,dis(q[i%l+1],q[j])); } return res;}int main(){ scanf("%d",&n); for(int i=1;i<=n;i++)scanf("%d%d",&p[i].x,&p[i].y); graham(); printf("%d\n",calliper()); return 0;}