## Topic

Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill between farmers and their cows. For simplicity, the world will be represented as a two-dimensional plane, where each farm is located at a pair of integer coordinates (x,y), each having a value in the range -10,000 ... 10,000. No two farms share the same pair of coordinates.

Even though Bessie travels directly in a straight line between pairs of farms, the distance between some farms can be quite large, so she wants to bring a suitcase full of hay with her so she has enough food to eat on each leg of her journey. Since Bessie refills her suitcase at every farm she visits, she wants to determine the maximum possible distance she might need to travel so she knows the size of suitcase she must bring.Help Bessie by computing the maximum distance among all pairs of farms.

### Input

• Line 1: A single integer, N

• Lines 2..N+1: Two space-separated integers x and y specifying coordinate of each farm

### Output

• Line 1: A single integer that is the squared distance between the pair of farms that are farthest apart from each other.

4

0 0

0 1

1 1

1 0

2

### Hint

Farm 1 (0, 0) and farm 3 (1, 1) have the longest distance (square root of 2)

The main idea of the topic ： Give the coordinates of some farms , Find the square of the linear distance between the farthest farms .

First, the convex hull is obtained by scanning , Rotate the chuck to get the maximum value

``````
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
#define MAX 50010
#define INF 1000000000
#define rg register
{
int x=0,t=1;char ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
if(ch=='-'){t=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
return x*t;
}
struct Node
{
int x,y;
}p[MAX],p0,S[MAX];
int n,top,T;
inline bool cmp(Node a,Node b)
{
rg double A=atan2(a.y-p0.y,a.x-p0.x);
rg double B=atan2(b.y-p0.y,b.x-p0.x);
if(A!=B)return A<B;
else return a.x<b.x;
}
inline long long chaji(int x1,int y1,int x2,int y2)// Calculate the cross product
{
return (1LL*x1*y2-1LL*x2*y1);
}
inline long long Compare(Node a,Node b,Node c)// Calculate the vector
{
return chaji((b.x-a.x),(b.y-a.y),(c.x-a.x),(c.y-a.y));
}
inline void Find()// Looking for convex hull
{
p0=(Node){INF,INF};
rg int k=0;
for(rg int i=0;i<n;++i)// Find the bottom point
if(p0.y>p[i].y||(p0.y==p[i].y&&p0.x>p[i].x))
p0=p[i],k=i;
swap(p[k],p);
sort(&p,&p[n],cmp);// About sorting the bottom points
S=p;S=p;
top=1;// To the top of the stack
for(rg int i=2;i<n;)// Find the convex hull
{
if(top&&Compare(S[top-1],p[i],S[top])>=0) top--;
else S[++top]=p[i++];
}
}
inline long long Dis(Node a,Node b)// Calculate the sum of the squares of the distances between two points
{
return 1LL*(a.x-b.x)*(a.x-b.x)+1LL*(a.y-b.y)*(a.y-b.y);
}
long long GetMax()// Find the diameter
{
rg long long re=0;
if(top==1)// There are only two points
return Dis(S,S);
S[++top]=S;// Put the first point at the end
int j=2;
for(int i=0;i<top;++i)// Enumerate edges
{
while(Compare(S[i],S[i+1],S[j])<Compare(S[i],S[i+1],S[j+1]))
j=(j+1)%top;
re=max(re,max(Dis(S[i],S[j]),Dis(S[i+1],S[j])));
}
return re;
}
int main()
{
for(int i=0;i<n;++i)
{
}
long long ans=INF,ss;
Find();
ans=GetMax();
cout<<ans<<endl;
return 0;
}
``````

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