简单几何(凸包) POJ 2187 Beauty Contest

题目传送门

题意：求两点的距离平方的最大值

分析：凸包模板题

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/25 9:31:11
* File Name     :A.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 5e4 + 10;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-10;
int dcmp(double x)  {       //三态函数，减少精度问题
    if (fabs (x) < EPS) return 0;
    else    return x < 0 ? -1 : 1;
}
struct Point    {       //点的定义
    double x, y;
    Point (double x=0, double y=0) : x (x), y (y) {}
    Point operator + (const Point &r) const {       //向量加法
        return Point (x + r.x, y + r.y);
    }
    Point operator - (const Point &r) const {       //向量减法
        return Point (x - r.x, y - r.y);
    }
    Point operator * (double p)  {       //向量乘以标量
        return Point (x * p, y * p);
    }
    Point operator / (double p)  {       //向量除以标量
        return Point (x / p, y / p);
    }
    bool operator < (const Point &r) const {       //点的坐标排序
        return x < r.x || (x == r.x && y < r.y);
    }
    bool operator == (const Point &r) const {       //判断同一个点
        return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;
    }
};
typedef Point Vector;       //向量的定义
Point read_point(void)   {      //点的读入
    double x, y;
    scanf ("%lf%lf", &x, &y);
    return Point (x, y);
}
double polar_angle(Vector A)  {     //向量极角
    return atan2 (A.y, A.x);
}
double dot(Vector A, Vector B)  {       //向量点积
    return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B)    {       //向量叉积
    return A.x * B.y - A.y * B.x;
}
double length(Vector A) {       //向量长度，点积
    return sqrt (dot (A, A));
}
double angle(Vector A, Vector B)    {       //向量转角，逆时针，点积
    return acos (dot (A, B) / length (A) / length (B));
}
double area_triangle(Point a, Point b, Point c) {       //三角形面积，叉积
    return fabs (cross (b - a, c - a)) / 2.0;
}
Vector rotate(Vector A, double rad) {       //向量旋转，逆时针
    return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
}
Vector nomal(Vector A)  {       //向量的单位法向量
    double len = length (A);
    return Vector (-A.y / len, A.x / len);
}
Point point_inter(Point p, Vector V, Point q, Vector W)    {        //两直线交点，参数方程
    Vector U = p - q;
    double t = cross (W, U) / cross (V, W);
    return p + V * t;
} 

vector<Point> convex_hull(vector<Point> &P) {
    sort (P.begin (), P.end ());
    int n = P.size (), k = 0;
    vector<Point> ret (n * 2);
    for (int i=0; i<n; ++i) {
        while (k > 1 && cross (ret[k-1] - ret[k-2], P[i] - ret[k-1]) <= 0)  k--;
        ret[k++] = P[i];
    }
    for (int i=n-2, t=k; i>=0; --i)  {
        while (k > t && cross (ret[k-1] - ret[k-2], P[i] - ret[k-1]) <= 0)  k--;
        ret[k++] = P[i];
    }
    ret.resize (k-1);
    return ret;
}

vector<Point> p;

int main(void)    {
    int n;
    while (scanf ("%d", &n) == 1)   {
        p.clear ();
        for (int i=0; i<n; ++i) {
            p.push_back (read_point ());
        }
        vector<Point> qs = convex_hull (p);
        double ans = 0;
        for (int i=0; i<qs.size (); ++i)    {
            for (int j=0; j<i; ++j)    {
                ans = max (ans, dot (qs[i] - qs[j], qs[i] - qs[j]));
            }
        }
        printf ("%.0f\n", ans);
    }    

    return 0;
}