Analyzing the C ++ implementation point quadrilateral, circle inner

Code Description:

1. The point of a straight line relationship, determining a point is within the quadrilateral;

The distance between the point and the circle center point, determining a point is within a circle;

3. Description opencv display; (Of course, if the mask opencv, can easily determine whether the point inside a circle or a quadrangle)

#include <iostream>
#include <opencv2\opencv.hpp>

using namespace std;
using namespace cv;

struct pointLoc
{
	float x;
	float y;

};

//创建随机数据
void randData(int num, vector<pointLoc> &randomData)
{
	randomData.clear();
	pointLoc p;
	srand((unsigned)time(NULL));
	for (int n = 0; n < num; n++)
	{
		p.x = (int)100 * rand() / (RAND_MAX + 1);
		p.y = (int)200 * rand() / (RAND_MAX + 1);
		randomData.push_back(p);
	}
}

// 定义直线参数结构体  
struct LinePara
{
	float a;
	float b;
	float c;//变量满足等式ax + by + c = 0
};

// 获取直线参数  
void getLinePara(pointLoc p1, pointLoc p2, LinePara & LP)
{
	LP.a = p2.y - p1.y;
	LP.b = p1.x - p2.x;
	LP.c = p1.y * (p2.x - p1.x) - p1.x * (p2.y - p1.y);
}

/*
假如有第三个点p3(x3,y3)则判断p3与直线的位置关系可以按一下算法:
将p3(x3,y3)带入到直线AX+BY+C=0{A=y2-y1;B=x1-x2;C=y1*(x2-x1)-x1*(y2-y1)}中
如果A*x3+B*y3+C=0则表示点p3(x3,y3)在直线上。
如果A*x3+B*y3+C>0则表示点p3(x3,y3)在直线右侧。
如果A*x3+B*y3+C<0则表示点p3(x3,y3)在直线左侧。
*/
float eval(LinePara LP, pointLoc p)
{
	return (LP.a * p.x + LP.b * p.y + LP.c);
}

int main()
{
	vector<pointLoc> randomData;
	randData(50, randomData);
	
	/*自定义四边形的四个顶点
	a—b
	|  |
	d—c
	*/
	pointLoc p_a, p_b, p_c, p_d;
	p_a.x = 20;
	p_a.y = 20;
	p_b.x = 80;
	p_b.y = 20;
	p_d.x = 19;
	p_d.y = 80;
	p_c.x = 81;
	p_c.y = 81;

	LinePara line_a2b;
	getLinePara(p_a, p_b, line_a2b);
	LinePara line_b2c;
	getLinePara(p_b, p_c, line_b2c);
	LinePara line_c2d;
	getLinePara(p_c, p_d, line_c2d);
	LinePara line_d2a;
	getLinePara(p_d, p_a, line_d2a);

	//求在四边形内的点
	vector<pointLoc> pInPolygon;
	pInPolygon.clear();
	for (int m = 0; m < randomData.size(); m++)
	{
		float df = eval(line_a2b, randomData[m]);
		if (df > 0)
			continue;

		df = eval(line_b2c, randomData[m]);
		if (df > 0)
			continue;

		df = eval(line_c2d, randomData[m]);
		if (df > 0)
			continue;

		df = eval(line_d2a, randomData[m]);
		if (df > 0)
			continue;

		pInPolygon.push_back(randomData[m]);
	}

	//求在圆内的点
	Point centerP(50, 150);
	int raduis = 30;
	vector<pointLoc> pInCircle;
	pInCircle.clear();
	for (int m = 0; m < randomData.size(); m++)
	{
		if (sqrt(pow(randomData[m].x - centerP.x, 2) + pow(randomData[m].y - centerP.y, 2)) > raduis)
			continue;

		pInCircle.push_back(randomData[m]);
	}


	//显示结果
	cv::Mat xx = cv::Mat::zeros(200, 100, CV_8UC3);

	line(xx, Point(p_a.x, p_a.y), Point(p_b.x, p_b.y), Scalar(0, 255, 0));
	line(xx, Point(p_b.x, p_b.y), Point(p_c.x, p_c.y), Scalar(0, 255, 0));
	line(xx, Point(p_c.x, p_c.y), Point(p_d.x, p_d.y), Scalar(0, 255, 0));
	line(xx, Point(p_d.x, p_d.y), Point(p_a.x, p_a.y), Scalar(0, 255, 0));
	
	circle(xx, centerP, raduis, Scalar(0, 255, 0));

	for (int m = 0; m < randomData.size(); m++)
	{
		xx.at<Vec3b>(randomData[m].y, randomData[m].x)[0] = 255;
		xx.at<Vec3b>(randomData[m].y, randomData[m].x)[1] = 0;
		xx.at<Vec3b>(randomData[m].y, randomData[m].x)[2] = 0;
	}

	for (int m = 0; m < pInPolygon.size(); m++)
	{
		xx.at<Vec3b>(pInPolygon[m].y, pInPolygon[m].x)[0] = 0;
		xx.at<Vec3b>(pInPolygon[m].y, pInPolygon[m].x)[1] = 0;
		xx.at<Vec3b>(pInPolygon[m].y, pInPolygon[m].x)[2] = 255;
	}
	for (int m = 0; m < pInCircle.size(); m++)
	{
		xx.at<Vec3b>(pInCircle[m].y, pInCircle[m].x)[0] = 0;
		xx.at<Vec3b>(pInCircle[m].y, pInCircle[m].x)[1] = 255;
		xx.at<Vec3b>(pInCircle[m].y, pInCircle[m].x)[2] = 255;
	}
}