PKU 2187 Beauty Contest 凸包+旋转卡壳

 

Beauty Contest

Description

Bessie, Farmer John's prize cow, has just won first placein 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 inorder to spread goodwill between farmers and their cows. For simplicity, theworld will be represented as a two-dimensional plane, where each farm islocated at a pair of integer coordinates (x,y), each having a value in therange -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 asuitcase full of hay with her so she has enough food to eat on each leg of herjourney. Since Bessie refills her suitcase at every farm she visits, she wantsto determine the maximum possible distance she might need to travel so sheknows the size of suitcase she must bring.Help Bessie by computing the maximumdistance among all pairs of farms. 

Input

* Line 1: A singleinteger, N 

* Lines 2..N+1: Two space-separated integers x and y specifying coordinate ofeach farm 

Output

* Line 1: A singleinteger that is the squared distance between the pair of farms that arefarthest apart from each other. 

Sample Input

4

0 0

0 1

1 1

1 0

Sample Output

2

 

旋转卡壳+凸包

#include<iostream>#include<algorithm>#include<cmath>#include<stdio.h>#include<cstdlib>using namespace std;const int N=50005;const double esp=1.0e-8;typedef struct{double x,y;}point;point p[N],t[N];int dblcmp(double x){if(fabs(x)<esp)return 0;return (x>0)?1:-1;}bool cmp(const point& e1,const point& e2){return e1.y<e2.y||(e1.y==e2.y&&e1.x<e2.x);}double distances(point a,point b){return sqrt( (b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y) );}double crossleft(point a,point b,point c){return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);}double rotating_calipers(int m){ int q=1; double ans=0.0; t[m]=t[0]; for(int p=0;p<m;p++) { while(crossleft(t[p],t[p+1],t[q+1])>crossleft(t[p],t[p+1],t[q])) q=(q+1)%m; ans=max(ans,max(distances(t[p],t[q]),distances(t[p+1],t[q+1]))); } return ans*ans; }int Jarvis(int n){int k,top=0;t[top++]=p[0];t[top++]=p[1];for(int i=2;i<n;i++){while(top>1&&dblcmp( crossleft(t[top-2],t[top-1],p[i]) )<=0)top--;t[top++]=p[i];}k=top;t[top++]=p[n-2];for(int i=n-3;i>=0;i--){while(top>k&&dblcmp( crossleft(t[top-2],t[top-1],p[i]) )<=0)top--;t[top++]=p[i];}return top-1;}int main(){//freopen("1.txt","r",stdin);int n,m;scanf("%d",&n);for(int i=0;i<n;i++)scanf("%lf %lf",&p[i].x,&p[i].y);sort(p,p+n,cmp);m=Jarvis(n);int ans=rotating_calipers(m);printf("%d/n",ans);return 0;}