[人工智能]OpenCV 角点排序 demo（C++版）

前言

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??OpenCV 中确定矩形 ROI 区域后，可获取矩形区域中四个角点，若进行 solvePnP 测距，角点要一一对应，故需对其进行排序。
??其次，若矩形发生大于90度的旋转（无论顺时针还是逆时针），写死的 “目标实际点位” 便会与 “测得的角点” 发生错位，影响 solvePnP 测距，所以还需考虑矩形某一特定方向上是长轴还是短轴，下面给出角点排序的 demo 与大家交流。

一、C++ 角点排序 demo

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	//********************** 坐标点排序 **********************
	Point2f verticess[4];
	Point2f tempPoint;
	verticess[0] = Point2f(50, 10);
	verticess[1] = Point2f(100, 10);
	verticess[2] = Point2f(100, 30);
	verticess[3] = Point2f(50, 30);

	double longSide, shortSide;    //定义长短边

	// 按x轴坐标从小到大排序
	for (int i = 0; i < 3; i++)
	{
		for (int j = i + 1; j < 4; j++)
		{
			if (verticess[i].x > verticess[j].x)
			{
				tempPoint = verticess[i];
				verticess[i] = verticess[j];
				verticess[j] = tempPoint;
			}
		}
	}

	// 按纵坐标排序，依次排为()： 0左上角   1左下角   2右上角   3右下角
	if (verticess[0].y > verticess[1].y)
	{
		tempPoint = verticess[0];
		verticess[0] = verticess[1];
		verticess[1] = tempPoint;
	}
	if (verticess[2].y > verticess[3].y)
	{
		tempPoint = verticess[2];
		verticess[2] = verticess[3];
		verticess[3] = tempPoint;
	}

	// 输出排序后结果
	for (int i = 0; i < 4; i++)
	{
		cout << "verticess:  " << verticess[i] << endl;

	}

	// 存入Point2f类容器中，以左下角为第一个点，逆时针方向存入
	urelpoints.clear();
	urelpoints.push_back(verticess[1]);
	urelpoints.push_back(verticess[3]);
	urelpoints.push_back(verticess[2]);
	urelpoints.push_back(verticess[0]);

	// 计算长短边，适应两种矩形
	longSide = sqrt(pow(verticess[2].x - verticess[0].x, 2) + pow(verticess[2].y - verticess[0].y, 2));
	shortSide = sqrt(pow(verticess[1].x - verticess[0].x, 2) + pow(verticess[1].y - verticess[0].y, 2));

	cout << "longSide:  " << longSide << endl;
	cout << "shortSide:  " << shortSide << endl;


	// 输出目标容器结果
	for (int i = 0; i < 4; i++)
	{
		cout << "urelpoints :  " << urelpoints[i] << endl;
	}

	
	if (shortSide > longSide)     // 短轴大于长轴，则为竖直方向为长轴的矩形
	{
		// 采用objA

	}
	else 
	{
		// 采用objB

	}

总结

??此博客为 C++版 OpenCV 角点排序，Python版参见本分类其他文章。