[POJ 2187] Beauty Contest

Beauty Contest

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 28659		Accepted: 8898

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

涨知识了、旋转卡(qia)壳
前面写凸包都是从1开始的、这里改为从0开始、方便一点
附暴力程序、或许是因为数据弱了、所以旋转卡壳算法再本题的时间上没有太大优化。

#include<iostream>

#include<algorithm>

#include<cmath>

#include<cstdio>

using namespace std;

#define N 100010



struct Point

{

    int x,y;

    Point(){}

    Point (int x,int y):x(x),y(y){

    }

    Point operator -(Point p){

        return Point(x-p.x,y-p.y);

    }

    int operator * (Point p){

        return x*p.x+y*p.y;

    }

    int operator ^ (Point p){

        return x*p.y-y*p.x;

    }

    bool operator < (const Point &p)const

    {

        if(x!=p.x) return x<p.x;

        return y<p.y;

    }

};



int n;

int top;

Point p[N],q[N];



void Graham()

{

    int i;

    sort(p,p+n);

    top=0;

    for(i=0;i<n;i++)

    {

        while(top>1 && ((q[top-1]-q[top-2])^(p[i]-q[top-2]))<=0)

            top--;

        q[top++]=p[i];

    }

    int t=top;

    for(i=n-1;i>=0;i--)

    {

        while(top>t && ((q[top-1]-q[top-2])^(p[i]-q[top-2]))<=0)

            top--;

        q[top++]=p[i];

    }

    top-=1;

}

int dis(Point a,Point b)

{

    return (a-b)*(a-b);

}

void solve()

{

    int i,j,ans=-1;

    for(i=0;i<top-1;i++)

    {

        for(j=i+1;j<top;j++)

        {

            ans=max(ans,dis(q[i],q[j]));

        }

    }

    cout<<ans<<endl;

}

void Rotate_Caliper()

{

    Point v;

    int ans=0,cur=1;

    for(int i=0;i<top;i++)

    {

        v=q[i]-q[(i+1)%top];

        while((v^(q[(cur+1)%top]-q[cur]))<0)

            cur=(cur+1)%top;

        ans=max(ans,max(dis(q[i],q[cur]),dis(q[(i+1)%top],q[(cur+1)%top])));

    }

    cout<<ans<<endl;

}

int main()

{

    while(scanf("%d",&n)!=EOF)

    {

        for(int i=0;i<n;i++)

        {

            scanf("%d%d",&p[i].x,&p[i].y);

        }

        Graham();

        //暴力：solve();

        Rotate_Caliper();

    }

    return 0;

}