rt，人都已经要似了qwq

就是归并，复杂度 $O(nlogn)$，大框架是没问题的，但是Wa了5个点QAQ。

马蜂良好：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define double long double
inline int read(){
	int rt=0;	bool kk=0;	char g=getchar();
	while(g<'0'||g>'9')	kk|=(g=='-'),g=getchar();
	while(g>='0'&&g<='9')	rt=(rt<<3)+(rt<<1)+g-'0',g=getchar();
	return (kk?-rt:rt);
}
int n,num;
double ans=1e18;
struct node{double x,y;}p[400005],ls[400005];
inline bool cmp(const node &A,const node &B){return A.x<B.x;}
inline double dis(const node &A,const node &B){return sqrt(pow(A.x-B.x,2)+pow(A.y-B.y,2));}
inline void work(int L,int R)
{
	if(L==R)	return;
	int mid=(L+R)>>1;	node now=p[mid];
	work(L,mid);	work(mid+1,R);
	int l=L,r=mid+1;	num=0;
	while(l<=mid&&r<=R)	ls[++num]=(p[l].y<p[r].y?p[l++]:p[r++]);
	while(l<=mid)	ls[++num]=p[l++];
	while(r<=R)		ls[++num]=p[r++];
	num=0;
	for(int i=L;i<=R;i++)
	{
		p[i]=ls[i-L+1];
		if(abs(p[i].x-now.x)<ans)	ls[++num]=p[i];
	}
	int j=1;
	for(int i=1;i<num;i++)
	{
		j=max(i,j);
		while(j<num&&ls[j+1].y-ls[i].y<ans)
			j++,ans=min(ans,dis(ls[i],ls[j]));
	}
}
int main()
{
	n=read();
	for(int i=1,x,y;i<=n;i++)	x=read(),y=read(),p[i]={x,y};
	sort(p+1,p+1+n,cmp);
	work(1,n);
	printf("%.0Lf",ans*ans);
	return 0;
}