POJ 2187 Beauty Contest （凸包+旋转卡壳）

解题思路：很显然需要求一个凸包，距离最远的点显然凸包上，因为这道题的数据范围比较小，所以可以平方枚举，当然更好的做法是旋转卡壳法。

AC代码：

/*
* @Author: wchhlbt
* @Last Modified time: 2017-09-07
*/

//#include <bits/stdc++.h>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <limits>
#include <climits>
#include <cstdio>

#define Fori(x) for(int i=0;i<x;i++)
#define Forj(x) for(int j=0;j<x;j++)
#define maxn 50007
#define inf 0x3f3f3f3f
#define ONES(x) __builtin_popcount(x)
#define _ << " " <<
using namespace std;

typedef long long ll ;
const double eps =1e-8;
const int mod = 1000000007;
const double PI = acos(-1.0);
int dx[5] = {0,0,1,-1,0};
int dy[5] = {1,-1,0,0,0};
inline int read(){ int cnt; scanf("%d",&cnt); return cnt;}

typedef struct Point
{
double x,y;
Point(double x=0,double y=0):x(x),y(y){}
}Vector;

Vector operator + (const Vector & A,const Vector & B) { return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (const Point & A,const Point & B) { return Vector(A.x-B.x,A.y-B.y);}
Vector operator * (const Vector & A,double p) { return Vector(A.x*p,A.y*p);}
Vector operator / (const Vector & A,double p) { return Vector(A.x/p,A.y/p);}

bool operator < (const Point &A, const Point &B)
{
return A.x<B.x||(A.x==B.x&&A.y<B.y);
}

int dcmp(double x)
{
if(fabs(x)<eps) return 0;
return x<0? -1 : 1 ;
}

bool operator == (const Point &A,const Point &B)
{
return dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)==0;
}

//向量点乘
double Dot(const Vector &A, const Vector &B) { return A.x*B.x+A.y*B.y; }

//求向量的长度或两点距离
double Length(const Vector &A){ return (Dot(A,A)); }

//返回两个向量夹角的弧度值
double Angle(const Vector &A, const Vector &B)
{
return acos(Dot(A,B)/Length(A)/Length(B));
}

//向量叉积。共起点的两个向量A,B,若A叉乘B大于0,则B在A的左边。
double Cross(const Vector &A, const Vector &B){ return A.x*B.y-A.y*B.x ;}

//用叉积求三个点组成的三角形的面积。面积为0表示三点共线。
double Area2(const Point &A,const Point &B,const Point &C ){return Cross(B-A,C-A) ;}

//向量绕起点旋转，rad为逆时针旋转的弧度。
//Rotate(A,-rad)表示将A顺时针旋转rad
Vector Rotate(const Vector &A,double rad)
{
return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}

//计算A向量的单位法向量，即左转90度，将单位归一化
//调用前确保A向量不是零向量
Vector Normal(Vector A)
{
double L=Length(A);
return Vector(-A.y/L,A.x/L);
}

int ConvexHull (Point *p,int n ,Point *ch)
{
sort(p,p+n); //先比较x坐标，再比较y坐标
//上凸包
int m=0;
for(int i=0;i<n;i++)
{
while(m>1&&dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]))<=0) m--;
ch[m++]=p[i];
}
//下凸包
int k=m;
for(int i=n-2;i>=0;i--)
{
while(m>k&&dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]))<=0) m--;
ch[m++]=p[i];
}
if(n>1) m--;
return m;
}
//旋转卡壳法 求凸包直径p 凸包点集
//利用三角形面积法来判断到一条边最远的点是哪个点
//如果想要求出对踵点，可以开一个结构体维护到每个点最远的那个点的编号和两者距离
double RotatingCaliper(int m, Point *ch)
{
int q = 1;
ch[m] = ch[0];
double ans = 0.0;
for(int p = 0; p<m; p++){
while(fabs( Area2(ch[p],ch[p+1],ch[q+1]) ) > fabs( Area2(ch[p],ch[p+1],ch[q]) )) q = (q+1)%m;
ans = max(ans,max(Length(ch[q+1]-ch[p]) , Length(ch[q+1]-ch[p+1])));
ans = max(ans,max(Length(ch[q]-ch[p]) , Length(ch[q]-ch[p+1])));
}
return ans;
}
Point p[maxn];
Point ans[maxn];

int main()
{
int n;
while(~scanf("%d",&n)){
for(int i = 0; i<n; i++){
int x = read();
int y = read();
p[i] = Point(x,y);
}
int cnt = ConvexHull(p,n,ans);
double res = RotatingCaliper(cnt,ans);
printf("%.0f\n",res);
}
return 0;
}