1 //二分判定 覆盖问题 BZOJ 1052
2 // 首先确定一个最小矩阵包围所有点，则最优正方形的一个角一定与矩形一个角重合。
3 // 然后枚举每个角，再解决子问题
4
5 #include <bits/stdc++.h>
6 using namespace std;
7 #define LL long long
8 typedef pair<int,int> pii;
9 const int inf = 1e9;
10 const int MOD =5010;
11 const int N =5010;
12 #define clc(a,b) memset(a,b,sizeof(a))
13 const double eps = 1e-8;
14 void fre() {freopen("in.txt","r",stdin);}
15 void freout() {freopen("out.txt","w",stdout);}
16 inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}
17
18 struct node{
19 int x[20010],y[20010];
20 int num;
21 }a,b;
22 int n,mid;
23
24 void solve2(node &a,int x1,int y1,int x2,int y2){
25 int cnt=0;
26 for(int i=0;i<a.num;i++){
27 if(a.x[i]<x1||a.x[i]>x2||a.y[i]>y2||a.y[i]<y1){
28 a.x[cnt]=a.x[i];
29 a.y[cnt]=a.y[i];
30 cnt++;
31 }
32 }
33 a.num=cnt;
34 }
35 void solve(node &a,int f){
36 int x1=inf,x2=-inf,y1=inf,y2=-inf;
37 for(int i=0;i<a.num;i++){
38 x1=min(x1,a.x[i]);
39 x2=max(x2,a.x[i]);
40 y1=min(y1,a.y[i]);
41 y2=max(y2,a.y[i]);
42 }
43 if(f==1) solve2(a,x1,y1,x1+mid,y1+mid);
44 if(f==2) solve2(a,x2-mid,y1,x2,y1+mid);
45 if(f==3) solve2(a,x1,y2-mid,x1+mid,y2);
46 if(f==4) solve2(a,x2-mid,y2-mid,x2,y2);
47 }
48 bool check(){
49 for(int s=1;s<=4;s++){
50 for(int t=1;t<=4;t++){
51 b.num=a.num;
52 for(int i=0;i<b.num;i++){
53 b.x[i]=a.x[i],b.y[i]=a.y[i];
54 }
55 solve(b,s),solve(b,t);
56 int x1=inf,x2=-inf,y1=inf,y2=-inf;
57 for(int i=0;i<b.num;i++){
58 x1=min(x1,b.x[i]);
59 x2=max(x2,b.x[i]);
60 y1=min(y1,b.y[i]);
61 y2=max(y2,b.y[i]);
62 }
63 if(x2-x1<=mid&&y2-y1<=mid) return true;
64 }
65 }
66 return false;
67 }
68 int main(){
69 n=read();
70 for(int i=0;i<n;i++) a.x[i]=read(),a.y[i]=read();
71 a.num=n;
72 int l=0,r=inf;
73 while(l<=r){
74 mid=(l+r)>>1;
75 if(check()) r=mid-1;
76 else l=mid+1;
77 }
78 printf("%d\n",l);
79 return 0;
80 }
