POJ 2187 Beauty Contest （求最远点对，凸包+旋转卡壳）

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 24283   Accepted: 7420 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)  Source USACO 2003 Fal

 

 

旋转卡壳学习链接：

http://blog.csdn.net/ACMaker

http://www.cnblogs.com/xdruid/archive/2012/07/01/2572303.html

 

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

struct Point
{
    int x,y;
    Point(int _x = 0, int _y = 0)
    {
        x = _x;
        y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x, y - b.y);
    }
    int operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    int operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
    void input()
    {
        scanf("%d%d",&x,&y);
    }
};
int dist2(Point a,Point b)
{
    return (a-b)*(a-b);
}
const int MAXN = 50010;
Point list[MAXN];
int Stack[MAXN],top;
bool _cmp(Point p1,Point p2)
{
    int tmp = (p1-list[0])^(p2-list[0]);
    if(tmp > 0)return true;
    else if(tmp == 0 && dist2(p1,list[0]) <= dist2(p2,list[0]))
        return true;
    else return false;
}
void Graham(int n)
{
    Point p0;
    int k = 0;
    p0 = list[0];
    for(int i = 1;i < n;i++)
        if(p0.y > list[i].y || (p0.y == list[i].y && p0.x > list[i].x))
        {
            p0 = list[i];
            k = i;
        }
    swap(list[0],list[k]);
    sort(list+1,list+n,_cmp);
    if(n == 1)
    {
        top = 1;
        Stack[0] = 0;
        return;
    }
    if(n == 2)
    {
        top = 2;
        Stack[0] = 0;
        Stack[1] = 1;
        return;
    }
    Stack[0] = 0;
    Stack[1] = 1;
    top = 2;
    for(int i = 2;i < n;i++)
    {
        while(top > 1 && ((list[Stack[top-1]]-list[Stack[top-2]])^(list[i]-list[Stack[top-2]])) <= 0 )
            top--;
        Stack[top++] = i;
    }
}
//旋转卡壳，求两点间距离平方的最大值
int rotating_calipers(Point p[],int n)
{
    int ans = 0;
    Point v;
    int cur = 1;
    for(int i = 0;i < n;i++)
    {
        v = p[i]-p[(i+1)%n];
        while((v^(p[(cur+1)%n]-p[cur])) < 0)
            cur = (cur+1)%n;
        //printf("%d %d\n",i,cur);
        ans = max(ans,max(dist2(p[i],p[cur]),dist2(p[(i+1)%n],p[(cur+1)%n])));
    }
    return ans;
}
Point p[MAXN];
int main()
{
    int n;
    while(scanf("%d",&n) == 1)
    {
        for(int i = 0;i < n;i++)
            list[i].input();
        Graham(n);
        for(int i = 0;i < top;i++)
            p[i] = list[Stack[i]];
        printf("%d\n",rotating_calipers(p,top));
    }
    return 0;
}