题意:给定平面上横纵坐标在-1e9~1e9内的20000个整数点的坐标，用三个大小相同边平行于坐标轴的正方形覆盖(在边界上的也算)，问正方形的边长最小为多少?(整数)
思路:构造一个覆盖所有点的矩形，正方形的端点即为矩形的一角，这样枚举四个角的两个正方形，二分最大长度，看剩下的点是否能被第三个正方形覆盖。
ps：构造矩形的思想是看了题解才想到的。。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
using namespace std;
#define rep(i,n) for((i) = 0;i < n;i++)
struct Point{
int x[], y[],top;
}a;
const int inf = ;
void cut(Point &a,int x1,int y1,int x2,int y2)
{
Point b;
b.top = a.top;
for(int k = ;k <= a.top;k++)
b.x[k] = a.x[k],b.y[k] = a.y[k];
a.top = ;
for(int i = ;i <= b.top;i++)
if(b.x[i] < x1 || b.x[i] > x2 || b.y[i] < y1 || b.y[i] > y2){
a.x[++a.top] = b.x[i];
a.y[a.top] = b.y[i];
}
}
void solve(Point &a,int id,int mid)
{
int x1 = inf ,x2 = -inf ,y1 = inf ,y2 = -inf;
for(int k = ;k <= a.top;k++){
x1 = min(x1,a.x[k]);x2 = max(x2,a.x[k]);
y1 = min(y1,a.y[k]);y2 = max(y2,a.y[k]);
}
if(id == )
cut(a,x1,y1,x1+mid,y1+mid);
if(id == )
cut(a,x1,y2-mid,x1+mid,y2);
if(id == )
cut(a,x2-mid,y1,x2,y1+mid);
if(id == )
cut(a,x2-mid,y2-mid,x2,y2);
}
bool judge(int mid)
{
Point b;
for(int i = ;i <= ;i++){
for(int j = ;j <= ;j++){
b.top = a.top;
for(int k = ;k <= b.top;k++)
b.x[k] = a.x[k],b.y[k] = a.y[k];
solve(b,i,mid);
solve(b,j,mid); //可以只是一个正方形；
int x1 = inf ,x2 = -inf ,y1 = inf ,y2 = -inf;
for(int k = ;k <= b.top;k++){
x1 = min(x1,b.x[k]);x2 = max(x2,b.x[k]);
y1 = min(y1,b.y[k]);y2 = max(y2,b.y[k]);
}
if(x2 - x1 <= mid && y2 - y1 <= mid) return true;
}
}
return false;
}
int main()
{
int n,xx,yy;
cin>>n;
a.top = n;
for(int i = ;i <= n;i++){
scanf("%d%d",&xx,&yy);
a.x[i] = xx;
a.y[i] = yy;
}
int l = ,r = inf,ans;
while(l <= r){
int mid = (l + r) >> ;
if(judge(mid))
ans = mid,r = mid - ;
else l = mid + ;
}
printf("%d\n",ans);
return ;
}