POJ2187(旋转卡壳)

Beauty Contest

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 35459		Accepted: 10978

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) 

 

利用旋转卡壳求最远点对的距离平方。

 //2016.10.2
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #define N 50005
 #define eps 1e-8
 
 using namespace std;
 
 int n;
 
 struct point
 {
     double x, y;
     point(){}
     point(double a, double b):x(a), y(b){}
     point operator-(point a){//向量减法
         return point(x-a.x, y-a.y);
     }
     double operator*(point a){//向量叉积
         return x*a.y-y*a.x;
     }
     bool operator<(const point a)const{
         if(fabs(x-a.x)<eps)return y<a.y;//浮点数的判等不能直接用‘==’直接比较
         return x<a.x;
     }
     double len2(){//向量模的平方
         return x*x+y*y;
     }
 }p[N];
 
 struct polygon
 {
     int n;
     point p[N];
 }pg;
 
 double cp(point o, point a, point b)//向量oa，ob叉积
 {
     return (a-o)*(b-o);
 }
 
 void Convex(int &n)//Graham扫描法
 {
     sort(p, p+n);
     int top, m;
     pg.p[] = p[]; pg.p[] = p[]; top = ;
     for(int i = ; i < n; i++)//从前往后扫
     {
         while(top> && cp(p[i], pg.p[top], pg.p[top-])>=)top--;
         pg.p[++top] = p[i];
     }
     m = top;
     pg.p[++top] = p[n-];
     for(int i = n-; i >= ; i--)//从后往前扫
     {
         while(top>m && cp(p[i], pg.p[top], pg.p[top-])>=)top--;
         pg.p[++top] = p[i];
     }
     pg.n = top;
 }
 
 int rotating_calipers()//旋转卡壳
 {
     int v = ;n = pg.n;
     double ans = ;
     pg.p[n] = pg.p[];
     for(int u = ; u < n; u++)//旋转
     {
         while(cp(pg.p[u],pg.p[u+],pg.p[v+])>cp(pg.p[u],pg.p[u+],pg.p[v]))v = (v+)%n;
         ans = max(ans, max((pg.p[u]-pg.p[v]).len2(), (pg.p[u+]-pg.p[v+]).len2()));
     }
     return ans;
 }
 
 int main()
 {
     int n;
     while(scanf("%d", &n)!=EOF && n)
     {
         for(int i = ; i < n; i++)
               scanf("%lf%lf", &p[i].x, &p[i].y);
         Convex(n);
         int ans = rotating_calipers();
         printf("%d\n", ans);
     }
 
     return ;
 }