C++中根据一个点推断距离该点某个单位处的点的方法

矫情吗；* 2022-05-30 01:22 174阅读 0赞

int main()
    {
    	//创建一个用于绘制图像的 空白图  
    	cv::Mat image = cv::Mat::zeros(1024, 19200, CV_8UC3);
    
    	//输入拟合点  
    	std::vector<cv::Point> points;
    	/*points.push_back(cv::Point(500, 158));
    	points.push_back(cv::Point(105, 98));
    	points.push_back(cv::Point(155, 160));
    	points.push_back(cv::Point(212, 220));
    	points.push_back(cv::Point(248, 260));
    	points.push_back(cv::Point(320, 300));
    	points.push_back(cv::Point(350, 360));
    	points.push_back(cv::Point(412, 400));*/
    	points.push_back(cv::Point(50, 500));
    	points.push_back(cv::Point(50, 300));
    	points.push_back(cv::Point(250, 300));
    	points.push_back(cv::Point(250, 500));
    
    	//将拟合点绘制到空白图上  
    	for (int i = 0; i < points.size(); i++)
    	{
    		cv::circle(image, points[i], 5, cv::Scalar(0, 0, 255), 2, 8, 0);
    	}
    
    	cv::Vec4f line_para;
    
    	cv::fitLine(points, line_para, cv::DIST_L2, 0, 1e-2, 1e-2);
    
    	std::cout << "line_para = " << line_para << std::endl;
    
    	//获取点斜式的点和斜率  
    	cv::Point point0;
    	point0.x = line_para[2];
    	point0.y = line_para[3];
    
    	double k = line_para[1] / line_para[0];
    	//计算直线的端点(y = k(x - x0) + y0)  
    	cv::Point point1, point2;
    	point1.x = 20;
    	point1.y = k*(20 - point0.x) + point0.y;
    	point2.x = 100;
    	point2.y = k*(100 - point0.x) + point0.y;
    
    	//计算距离point0 20个单位出的点的位置并画出该段线
    	int step = 20;
    	Point pt1, pt2;
    	//后step处的点
    	pt1.x = point1.x - step;
    	pt1.y = point1.y - k*step;
    	//前step处的点
    	pt2.x = point1.x + step;
    	pt2.y = point1.y + k*step;
    	//Rect rect(pt2.x, pt2.y, 40, 40);
    
    	cv::circle(image, point1, 5, cv::Scalar(0,255, 0), 2, 8, 0);
    	waitKey(0);
    	cv::circle(image,pt1, 5, cv::Scalar(0, 0, 255), 2, 8, 0);
    	waitKey(0);
    	cv::circle(image, pt2, 5, cv::Scalar(255, 0, 0), 2, 8, 0);
    	waitKey(0);
    	cv::line(image, point1, point2, cv::Scalar(0, 255, 0), 2, 8, 0);
    	
    	cv::imshow("image", image);
    	cv::waitKey(0);
    	return 0;
    }
    算法实现部分：

int step = 20;
    	Point pt1, pt2;
    	//后step处的点
    	pt1.x = point1.x - step;
    	pt1.y = point1.y - k*step;
    	//前step处的点
    	pt2.x = point1.x + step;
    	pt2.y = point1.y + k*step;

这里的step为距离的单位，即在本例子中就是距离point1点横坐标的距离为20个单位。-表示在该点的左边，+表示在该点的右边。

*该方法适用于所有直线表达式。*

图片：![Center][]

绿点为本例中要实验的点，红点距离该点x坐标-20个单位的点，蓝点距离该点x坐标+20个单位的点。

[Center]: /images/20220530/259d2403a6514223aae41f7a87973b2e.png