poj 2187(凸包+旋转卡壳)

Beauty Contest

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33122		Accepted: 10282

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) 

方法1：暴力枚举，因为在算凸包的时候栈保存的是凸包上的顶点，所以直接枚举顶点即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int N = 50005;
struct Point{
    int x,y;
}p[N],Stack[N];
int n;

int mult(Point a,Point b,Point c){
    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}
int dis(Point a,Point b){
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int cmp(Point a,Point b){
    if(mult(a,b,p[0])>0) return 1;
    if(mult(a,b,p[0])==0&&dis(b,p[0])>dis(a,p[0])) return 1;
    return 0;
}
int Graham(){
    sort(p+1,p+n,cmp);
    int top = 2;
    Stack[0]=p[0];
    Stack[1]=p[1];
    Stack[2]=p[2];
    for(int i=3;i<n;i++){
        while(top>=1&&mult(p[i],Stack[top],Stack[top-1])>=0){
            top--;
        }
        Stack[++top]=p[i];
    }
    return top;
}
int main()
{
    while(scanf("%d",&n)!=EOF){
        for(int i=0;i<n;i++){
            scanf("%d%d",&p[i].x,&p[i].y);
        }
        int k = 0;
        for(int i=1;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();
        int ans =0;
        for(int i=0;i<=top;i++){
            for(int j=i+1;j<=top;j++){
                if(ans<dis(Stack[i],Stack[j])){
                    ans = dis(Stack[i],Stack[j]);
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

方法2：旋转卡壳，由于自己还没完全弄懂，所以暂不做详细讲解。

参考博客： http://www.cnblogs.com/liyinggang/p/5431908.html

     http://www.cnblogs.com/DreamUp/archive/2010/09/16/1828131.html

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int N = 50005;
struct Point{
    int x,y;
}p[N],Stack[N];
int n;

int mult(Point a,Point b,Point c){
    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}
int dis(Point a,Point b){
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int cmp(Point a,Point b){
    if(mult(a,b,p[0])>0) return 1;
    if(mult(a,b,p[0])==0&&dis(b,p[0])>dis(a,p[0])) return 1;
    return 0;
}
int Graham(){
    sort(p+1,p+n,cmp);
    int top = 2;
    Stack[0]=p[0];
    Stack[1]=p[1];
    Stack[2]=p[2];
    for(int i=3;i<n;i++){
        while(top>=1&&mult(p[i],Stack[top],Stack[top-1])>=0){
            top--;
        }
        Stack[++top]=p[i];
    }
    return top;
}
int rotating_calipers(int top)//旋转卡壳
{
    int q=1,ans=0,p;
    Stack[++top]=Stack[0]; ///将最后一个点和第一点的边也要算进去,于是多加一个点表示第一个点
    for(p=0; p<top; p++)
    {
　　　　 ///找到以Stack[p] 和 Stack[p+1]这两个点为底边的最大三角形 (叉积之积为三角形有向面积)(图1)
        while(mult(Stack[p],Stack[p+1],Stack[q+1])>mult(Stack[p],Stack[p+1],Stack[q+1]))
            q=(q+1)%top;
　　　　 ///在底边的两个点中选择与顶点距离大的
        ans=max(ans,max(dis(Stack[p],Stack[q]),dis(Stack[p+1],Stack[q])));
    }
    return ans;
}
int main()
{
    while(scanf("%d",&n)!=EOF){
        for(int i=0;i<n;i++){
            scanf("%d%d",&p[i].x,&p[i].y);
        }
        int k = 0;
        for(int i=1;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]);
        //printf("%d %d\n",p[0].x,p[0].y);
        int top = Graham();
        int ans =rotating_calipers(top);
        printf("%d\n",ans);
    }
    return 0;
}