hdu-2187 Beauty Contest（旋转卡壳求平面最远点对）

题目链接：点击打开链接

Beauty Contest

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 33134 Accepted: 10289

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

40 00 11 11 0

Sample Output

2

题意：求平面最远点对，也就是凸包直径

思路：旋转卡壳求最远点对模板。 关于旋转卡壳 详细内容请看：点击打开链接

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define N 50500#define PI 4*atan(1.0)#define eps 1e-8struct Node{    int x,y;} p[N],stack[N];int mulit(Node a,Node b,Node c)//向量ab与向量ac的叉乘{    return ((b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x));}int dist(Node s,Node e)//s点与e点的距离的平方（此处如果直接开根怕影响在传递的时候影响精度）{    return (s.x-e.x)*(s.x-e.x)+(s.y-e.y)*(s.y-e.y);}int cmp(Node a,Node b)//设<p1,p2,...pm>为对其余点按以p0为中心的极角逆时针排序所得的点集（如果有多个点有相同的极角，除了距p0最远的点外全部移除){    if(mulit(p[0],a,b)>0)//向量p[0]a在向量p[0]b顺时针方向,返回1        return 1;    if(mulit(p[0],b,a)==0&&(dist(p[0],b)-dist(p[0],a)>eps))//p[0]a与p[0]b共线，而且b点距离更远        return 1;    return 0;}int Graham(int n){    int top=2;    sort(p+1,p+n,cmp);    stack[0]=p[0];    stack[1]=p[1];    stack[2]=p[2];    for(int i=3; i<n; i++)    {        while(top>=1&&mulit(stack[top-1],p[i],stack[top])>=0)            top--;        stack[++top]=p[i];    }    return top;}int qiake(int top){    int q=1,ans=0;    stack[top+1]=stack[0];    for(int p=0; p<=top; p++)    {        while(mulit(stack[p+1],stack[q],stack[p])<mulit(stack[p+1],stack[q+1],stack[p]))            q=(q+1)%top;        ans=max(ans,max(dist(stack[p],stack[q]),dist(stack[p+1],stack[q+1])));    }    return ans;}int main(){    int T,n,m;    while(~scanf("%d",&n)&&n)    {        for(int i=0; i<n; i++)            scanf("%d %d",&p[i].x,&p[i].y);        int k=0;        for(int i=0; i<n; i++)        {            if(p[k].y>p[i].y||(p[k].y==p[i].y&&p[k].x>p[i].x))                k=i;        }        swap(p[0],p[k]);        int top=Graham(n);        int ans=qiake(top);        printf("%d\n",ans);    }    return 0;}