图像形状骨架提取细化 C++ 代码_c哩

// Thin(Skeleton).cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。 // #include <iostream> #include <opencv2/opencv.hpp> //https://blog.csdn.net/hehedadaq/article/details/80303218 //https://www.cnblogs.com/xianglan/archive/2011/01/01/1923779.html #define OPT 1 void thin(const cv::Mat &src, cv::Mat &dst) { int rows = src.rows; int cols = src.cols; // 映射表 uchar array[] = { 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0 }; for (int r = 0; r < rows; r++) { for (int c = 0; c < cols; c++) { uchar v = src.at<uchar>(r, c); if (v == 0) { //uchar a[9] = { 1 }; ///< 注意，这样只初始化了第一项 uchar a[9] = { 1, 1, 1, 1, 1, 1, 1, 1, 1 }; for (int k = 0; k < 3; k++) { for (int l = 0; l < 3; l++) { // 如果 3 * 3 矩阵的点不在边界且这些值为 0，也就是黑色的点 int posy = r - 1 + k; int posx = c - 1 + l; if (-1 < (posy) && (posy) < rows && -1 < (posx) && (posx) < cols && dst.at<uchar>(posy, posx) == 0) { a[k * 3 + l] = 0; } } } /// @test //for (int i = 0; i < 9; i++) //{ // std::cout << a[i] << " "; //} //std::cout << std::endl; // 然后根据 array 表，对 dst 的那一点进行赋值 /// 注意没有 a[4] int sum = a[0] * 1 + a[1] * 2 + a[2] * 4 + a[3] * 8 + a[5] * 16 + a[6] * 32 + a[7] * 64 + a[8] * 128; //std::cout << "sum: " << sum << std::endl; dst.at<uchar>(r, c) = array[sum] * 255; } } } } void thinH(const cv::Mat &src, cv::Mat &dst) { int rows = src.rows; int cols = src.cols; // 映射表 uchar array[] = { 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0 }; uchar NEXT = 1; for (int j = 0; j < cols; j++) { for (int i = 0; i < rows; i++) { if (NEXT == 0) { NEXT = 1; } else { int M = 1; if (i > 0 && i < rows - 1) M = src.at<uchar>(i - 1, j) + src.at<uchar>(i, j) + src.at<uchar>(i + 1, j); if (src.at<uchar>(i, j) == 0 && M != 0) { uchar a[9] = { 0, 0, 0, 0, 0, 0, 0, 0, 0 }; for (int k = 0; k < 3; k++) { for (int l = 0; l < 3; l++) { // 如果 3 * 3 矩阵的点不在边界且这些值为 0，也就是黑色的点 int posy = i - 1 + k; int posx = j - 1 + l; if (-1 < (posy) && (posy) < rows && -1 < (posx) && (posx) < cols && dst.at<uchar>(posy, posx) == 255) { a[k * 3 + l] = 1; } } } int sum = a[0] * 1 + a[1] * 2 + a[2] * 4 + a[3] * 8 + a[5] * 16 + a[6] * 32 + a[7] * 64 + a[8] * 128; dst.at<uchar>(i, j) = array[sum] * 255; if (array[sum] == 1) NEXT = 0; } } } } } void thinV(const cv::Mat &src, cv::Mat &dst) { int rows = src.rows; int cols = src.cols; // 映射表 uchar array[] = { 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0 }; uchar NEXT = 1; for (int i = 0; i < rows; i++) { for (int j = 0; j < cols; j++) { if (NEXT == 0) { NEXT = 1; } else { int M = 1; if (j > 0 && j <cols - 1) M = src.at<uchar>(i, j - 1) + src.at<uchar>(i, j) + src.at<uchar>(i, j + 1); if (src.at<uchar>(i, j) == 0 && M != 0) { uchar a[9] = { 0, 0, 0, 0, 0, 0, 0, 0, 0 }; for (int k = 0; k < 3; k++) { for (int l = 0; l < 3; l++) { // 如果 3 * 3 矩阵的点不在边界且这些值为 0，也就是黑色的点 int posy = i - 1 + k; int posx = j - 1 + l; if (-1 < (posy) && (posy) < rows &&