【openCV】二值图像基础操作

最新推荐文章于 2025-06-09 13:54:11 发布

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最新推荐文章于 2025-06-09 13:54:11 发布 · 3.3k 阅读

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CC 4.0 BY-SA版权

文章标签：

#openCV #thin #thick #skeleton

实现了二值图像的Thin,Thicken和提取骨架的操作。

#include "opencv2/opencv.hpp"

#define HIT 1
#define MISS 0

using namespace cv;
using namespace std;

const int dir[9][2] = {
  
  {-1,-1},{-1,0},{-1,1},{0,-1},{0,0},{0,1},{1,-1},{1,0},{1,1}};

//定义了skeleton和convex hull操作用到的各八个structuring element
//structuring element可根据不同需要改变，这里只是一种
#define ELE_STR_SIZE 3

#define SKELETON_STR_ELE_NUM 8
const unsigned char skeleton_str_ele[SKELETON_STR_ELE_NUM][ELE_STR_SIZE][ELE_STR_SIZE] = {
	{
		{0,0,0},
		{2,1,2},
		{1,1,1}
	},

	{
		{2,0,0},
		{1,1,0},
		{2,1,2}
	},

	{
		{1,2,0},
		{1,1,0},
		{1,2,0}
	},

	{
		{2,1,2},
		{1,1,0},
		{2,0,0}
	},

	{
		{1,1,1},
		{2,1,2},
		{0,0,0}
	},

	{
		{2,1,2},
		{0,1,1},
		{0,0,2}
	},

	{
		{0,2,1},
		{0,1,1},
		{0,2,1}
	},

	{
		{0,0,2},
		{0,1,1},
		{2,1,2}
	}
};

#define CONVEX_HULL_STR_ELE_NUM 8
const unsigned char convex_hull_str_ele[CONVEX_HULL_STR_ELE_NUM][ELE_STR_SIZE][ELE_STR_SIZE] = {
	{
		{1,1,2},
		{1,0,2},
		{1,2,0}
	},

	{
		{2,1,1},
		{2,0,1},
		{0,2,1}
	},

	{
		{1,1,1},
		{2,0,1},
		{0,2,2}
	},

	{
		{0,2,2},
		{2,0,1},
		{1,1,1}
	},

	{
		{0,2,1},
		{2,0,1},
		{2,1,1}
	},

	{
		{1,2,0},
		{1,0,2},
		{1,1,2}
	},

	{
		{2,2,0},
		{1,0,2},
		{1,1,1}
	},

	{
		{1,1,1},
		{1,0,2},
		{2,2,0}
	}
};


unsigned char hit_and_miss(unsigned char src[][ELE_STR_SIZE],const unsigned char str_ele[][ELE_STR_SIZE]){
	//对一块区域的hit_and_miss判定
	for (int i = 0;i < ELE_STR_SIZE;i++){
		for (int j = 0;j < ELE_STR_SIZE;j++){
			if (str_ele[i][j] == 0 || str_ele[i][j] == 1){
				if (str_ele[i][j] != src[i][j]) return MISS;
			}
		}
	}
	return HIT;
}

void hit_and_miss(IplImage *src,IplImage **dst,const unsigned char str_ele[][ELE_STR_SIZE])
{
	//对一个二值图像的hit_and_miss操作
	unsigned char matrix[ELE_STR_SIZE][ELE_STR_SIZE];
	int i1,j1,nx,ny;
	CvSize size = cvGetSize(src);
	if ((*dst) != NULL) cvReleaseImage(dst); 
	(*dst) = cvCreateImage(cvGetSize(src), IPL_DEPTH_8U, 1);
	cvZero(*dst);

	for (int i = 0;i<size.height;i++){
		for (int j = 0;j<size.width;j++){
			
			i1 = 0;
			j1 = 0;
			for (int k = 0;k < 9;k++){
				//把自身及相邻共九个格子放入一个矩阵中
				nx = i + dir[k][0];
				ny = j + dir[k][1];
				if (nx<0 || nx>=size.height ||