c++ - c++顺时针排序2D点

我编写了一个程序，从12点开始以顺时针方式在图形上排列点，这样，包含这些点的 vector 就按该顺序排序。我正在使用atan2从12点开始获取 Angular ，然后根据象限进行调整。我试图找出错误的来源，因为它没有正确订购它们。因此，给定4个随机点(如照片中的随机点)，它应该在包含 vector 的顺序为P1，P2，P3，P4
这是我的代码:

//sort_points.cpp
#include <iostream>
#include <math.h>
#include <algorithm>
#include <vector>

using namespace std;

class Point
{
    public:
        double x;
        double y;
        Point(double xx, double yy) : x(xx), y(yy) {}
        ~Point();
        inline friend ostream& operator<<(ostream& output, const Point& point)
        {
            output << "[" << point.x << ", " << point.y <<"]";
            return output;
        }
};


Point::~Point() {;}


/* get quadrant from 12 o'clock*/
int get_quadrant (const Point& p)
{
    int result = 4; //origin

    if (p.x > 0 && p.y > 0)
        return 1;
    else if(p.x < 0 && p.y > 0)
        return 2;
    else if(p.x < 0 && p.y < 0)
        return 3;
    //else 4th quadrant
    return result;
}

double get_clockwise_angle(const Point& p)
{
    double angle = 0.0;
    int quadrant = get_quadrant(p);

    /*making sure the quadrants are correct*/
    cout << "Point: " << p << " is on the " << quadrant << " quadrant" << endl;

    /*add the appropriate pi/2 value based on the quadrant. (one of 0, pi/2, pi, 3pi/2)*/
    switch(quadrant)
    {
        case 1:
            angle = atan2(p.x,p.y) * 180/M_PI;
            break;
        case 2:
            angle = atan2(p.y, p.x)* 180/M_PI;
            angle += M_PI/2;
            break;
        case 3:
            angle = atan2(p.x,p.y)* 180/M_PI;
            angle += M_PI;
            break;
        case 4:
            angle = atan2(p.y, p.x)* 180/M_PI;
            angle += 3*M_PI/2;
            break;
    }
    return angle;
}
bool compare_points(const Point& a, const Point& b)
{
    return (get_clockwise_angle(a) < get_clockwise_angle(b));
}
int main(int argc, char const *argv[])
{
    std::vector <Point> points;
    points.push_back( Point( 1, 3 ) );
    points.push_back( Point( 2, 1 ) );
    points.push_back( Point( -3, 2 ) );
    points.push_back( Point( -1, -1 ) );

    cout << "\nBefore sorting" << endl;
    for (int i = 0; i < points.size(); ++i)
    {
        cout << points.at(i) << endl;
    }

    std::sort(points.begin(), points.end(),compare_points);

    cout << "\nAfter sorting" << endl;
    for (int i = 0; i < points.size(); ++i)
    {
        cout << points.at(i) << endl;
    }
    return 0;
}

最佳答案

您不需要调整。 atan2将为您提供从x轴正方向到-PI到PI范围内的逆时针 Angular 。

首先，为使起点为y轴正方向，让我为atan2设置参数，好像y轴的负方向为x轴的正方向，而x轴的正方向为y轴。

然后，这将使 Angular 逆时针旋转，因此取消 Angular 以颠倒顺序。

double get_clockwise_angle(const Point& p)
{
    double angle = 0.0;
    int quadrant = get_quadrant(p);

    /*making sure the quadrants are correct*/
    cout << "Point: " << p << " is on the " << quadrant << " quadrant" << endl;

    /*calculate angle and return it*/
    angle = -atan2(p.x,-p.y);
    return angle;
}

关于c++ - c++顺时针排序2D点，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/39187184/