POJ 2187 Beauty Contest（凸包）

Description

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.

Sample Input

4
0 0
0 1
1 1
1 0

Sample Output

2

Hint

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

题目大意：

给定平面内一些点，求最大的两点间直线距离的平方

解题思路：

凸包裸题，纯粹模板，感觉数据是随机出的，所以跑一遍凸包，然后两两枚举凸包上的点，求出最大距离平方即可，不会超时，不需要旋转卡壳（PS：弱也不会旋转卡壳。。。。

#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <map>
#include <set>
#include <bitset>
#include <ctime>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <list>

#define STD_REOPEN() freopen("../in.in","r",stdin)
#define STREAM_REOPEN fstream cin("../in.in")
#define INF 0x3f3f3f3f
#define _INF 63
#define eps 1e-8
#define MAX_V 100010
#define MAX_P 50010
#define MAX_E 10010
#define MAX 32000
#define MOD_P 3221225473
#define MOD 9901

using namespace std;

struct Point
{
int x,y;
Point(){}
Point(int a,int b):x(a),y(b){}
Point operator - (const Point a)const	//求向量
{
return Point(x-a.x,y-a.y);
}
int operator ^ (const Point a)const	//叉乘
{
return (x*a.y)-(y*a.x);
}
int operator & (const Point a)const	//两点间距离
{
return (x-a.x)*(x-a.x)+(y-a.y)*(y-a.y);
}
bool operator < (const Point a)const
{
if(y==a.y)
return x<a.x;
return y<a.y;
}
}s[MAX_P];

int GrahamScan(int n)
{
if(n==2)
return s[0]&s[1];
sort(s,s+n);
Point res[MAX_P];
int top=0;
for(int i=0;i<n;i++)
{
while(top>=2&&(Point(res[top-1]-res[top-2])^Point(s[i]-res[top-2]))<0)
top--;
res[top++]=s[i];
}
int len=top+1;
for(int i=n-2;i>=0;i--)
{
while(top>=len&&(Point(res[top-1]-res[top-2])^Point(s[i]-res[top-2]))<0)
top--;
res[top++]=s[i];
}
int ans=0;
for(int i=0;i<top;i++)
for(int j=i+1;j<top;j++)
ans=max(ans,res[i]&res[j]);
return ans;
}

int main()
{
//STD_REOPEN();
int n;
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
scanf("%d %d",&s[i].x,&s[i].y);
int ans=GrahamScan(n);
printf("%d\n",ans);
}

return 0;
}