• 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
    ", ans);
        }    
    
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Running-Time/p/4908467.html
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