大意: 平面上n个点每个点坐标为(x,0)或(0,y), 求任意两点距离平方最大值的最小值.
二分答案, 转化为判定最大值是否<=e, 按$x$排序后, 因为固定左端点, $y$绝对值的最大值是跟右端点单调的, 滑动一个长度平方不超过e的区间, 同时保证右端点$x$的绝对值不超过左端点, 这样对于左端点在$x$轴的情况一定是最优的, 同样再固定右端点倒序处理正半轴的情况.
#include <iostream>
#include <random>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif
int n;
pii a[N];
int Lmin[N], Lmax[N], Rmin[N], Rmax[N];
ll sqr(ll x) {return x*x;}
ll ans = 1e18;
int chk(ll e) {
if (ans<=e) return 1;
int now = 1;
ll ans = 1e18;
REP(i,1,n) {
if (a[i].x>0) break;
while (now<n&&sqr(a[now+1].x-a[i].x)<=e&&abs(a[now+1].x)<=abs(a[i].x)) ++now;
while (abs(a[now].x)>abs(a[i].x)) --now;
int U = -1e9, D = 1e9;
if (i>1) U=max(U,Lmax[i-1]),D=min(D,Lmin[i-1]);
if (now<n) U=max(U,Rmax[now+1]),D=min(D,Rmin[now+1]);
ans = min(ans, max(sqr(U-D),max(sqr(U),sqr(D))+max(sqr(a[i].x),sqr(a[now].x))));
}
now = n;
PER(i,1,n) {
if (a[i].x<0) break;
while (now>1&&sqr(a[now-1].x-a[i].x)<=e&&abs(a[now-1].x)<=abs(a[i].x)) --now;
while (abs(a[now].x)>abs(a[i].x)) ++now;
int U = -1e9, D = 1e9;
if (i<n) U=max(U,Rmax[i+1]),D=min(D,Rmin[i+1]);
if (now>1) U=max(U,Lmax[now-1]),D=min(D,Lmin[now-1]);
ans = min(ans, max(sqr(U-D),max(sqr(U),sqr(D))+max(sqr(a[i].x),sqr(a[now].x))));
}
return ans<=e;
}
int main() {
scanf("%d", &n);
REP(i,1,n) scanf("%d%d", &a[i].x,&a[i].y);
sort(a+1,a+1+n);
Lmin[1]=Lmax[1]=a[1].y;
REP(i,2,n) {
Lmin[i]=min(Lmin[i-1],a[i].y);
Lmax[i]=max(Lmax[i-1],a[i].y);
}
Rmin[n]=Rmax[n]=a[n].y;
PER(i,1,n-1) {
Rmin[i]=min(Rmin[i+1],a[i].y);
Rmax[i]=max(Rmax[i+1],a[i].y);
}
ll l = 0, r = min(sqr(Lmin[n]-Lmax[n]),sqr(a[1].x-a[n].x));
ans = r;
while (l<=r) {
if (chk(mid)) ans=mid,r=mid-1;
else l=mid+1;
}
printf("%lld\n", ans);
}