Harris点特征提取算子c++

嗯.....本来打算一两天写一个简单的算法，没想到Harris进行的这么顺利，很快就完成了，那今天就写两个吧。承接上一篇的Forstner算子，继续完成Harris算子，参考资料为：

张祖勋《数字摄影测量学》

c++代码如下：

#include 
#include
using namespace std;
using namespace cv;
vectorfeaturePoints;
int k = 2;//5*5的窗口
int ksize = 5;//5*5的极值检测窗口
void Harris(Mat src)
{
	Mat I = Mat::zeros(src.rows, src.cols, CV_64FC1);//存储各像素的兴趣值
	Mat Ixx = Mat::zeros(src.rows, src.cols, CV_64FC1);//M左上角（高斯滤波前）
	Mat Iyy = Mat::zeros(src.rows, src.cols, CV_64FC1);//M右下角
	Mat Ixy = Mat::zeros(src.rows, src.cols, CV_64FC1);//M副对角线
	double GaussKernel[5][5] = //定义高斯卷积核
	{
		0.00390625, 0.015625, 0.0234375, 0.015625, 0.00390625,
		0.015625, 0.0625, 0.09375, 0.0625, 0.015625,
		0.0234375, 0.09375, 0.140625, 0.09375, 0.0234375,
		0.015625, 0.0625, 0.09375, 0.0625, 0.015625,
		0.00390625, 0.015625, 0.0234375, 0.015625, 0.00390625,
	};
	//1.一阶差分
	for (int r = k; r < src.rows - k; r++)
	{
		for (int c = k; c < src.cols - k; c++)
		{
			double gx = 0, gy = 0;
			for (int i = -k; i <= k - 1; i++)
			{
				gx += src.at(r, c + i) - src.at(r, c + i + 1);//x方向梯度
				gy += src.at(r + i, c) - src.at(r + i + 1, c);//y方向梯度
			}
			Ixx.at(r, c) = gx * gx;
			Ixy.at(r, c) = gx * gy;
			Iyy.at(r, c) = gy * gy;
		}
	}
	for (int r = k; r < src.rows - 5; r++)
	{
		for (int c = k; c < src.cols - 5; c++)
		{
			double M11 = 0, M12 = 0, M22 = 0;
			for (int j = 0; j < 5; j++)
			{
				for (int i = 0; i < 5; i++)
				{
					//2.对梯度值进行高斯滤波
					M11 += Ixx.at(r + j, c + i) * GaussKernel[j][i];//高斯滤波后的M值
					M12 += Ixy.at(r + j, c + i) * GaussKernel[j][i];
					M22 += Iyy.at(r + j, c + i) * GaussKernel[j][i];
				}
			}
			double detM = M11 * M22 - M12 * M12;
			double trM = M11 + M22;
			double k = 0.04;
			//3.计算兴趣值
			I.at(r, c) = detM - k * trM * trM;//计算兴趣值
		}
	}
	//cout << "I:" << I << endl;
	//4.选取极值点
	for (int r = ksize; r < src.rows - ksize; r += ksize)
	{
		for (int c = ksize; c < src.cols - ksize; c += ksize)
		{
			double max = 0;
			int Flag = 0;
			Point2f point;
			for (int j = -ksize; j <= ksize; j++)
			{
				for (int i = -ksize; i <= ksize; i++)
				{
					double value = I.at(r + j, c + i);
					//局部非最大值抑制
					if (value > max)
					{
						max = value;
						point.x = c + i;
						point.y = r + j;
						Flag = 1;
					}
				}
			}
			if (Flag == 1)
			{
				featurePoints.push_back(point);
			}
		}
	}
	cout << "特征点个数：" << featurePoints.size() << endl;
}
void drawPoint(Mat img)
{
	for (int i = 0; i < featurePoints.size(); i++)
	{
		int centerx = featurePoints[i].x;
		int centery = featurePoints[i].y;
		circle(img, Point(centerx, centery), 4, Scalar(255, 0, 0), 1);
	}
}
int main()
{
	Mat src = imread("C:\\Users\\Administrator\\Desktop\\images\\标定板.png", IMREAD_GRAYSCALE);
	Harris(src);
	Mat img = imread("C:\\Users\\Administrator\\Desktop\\images\\标定板.png", IMREAD_COLOR);
	drawPoint(img);
	imwrite("D:\\Software\\VS2019\\source\\repos\\摄影测量与三维重建\\3.Harris点特征提取算子\\result\\Harris点特征提取算子.jpg", img);
	waitKey(0);
	return 0;
}

这个算法比较简单，也可能有前边两个做铺垫，这个明显写起来很顺手，也没遇到太大的阻碍，原图和结果如下：