二值图像边缘提取算法C语言实现[愿创黑客联盟Bugeyes]

二值图像边缘提取算法C语言实现[原创]

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定义：每个像素的取值均为0或1，称这样的图像为二值图像。

算法：检查所有像素，若该像素为物体上与背景接触的像素（四连通像素中既有背景像素又有物体像素），则为边界。

程序：

#define M 30
#define N 20

void edge(int image[M][N],int bianyuan[M][N])
{
   int i,j;
   int inner=1,outer=1;
   for (i=0;i      for(j=0;j            bianyuan[i][j]=0;
   for(i=1;i      for(j=1;j      {
         inner=1;/*假设该像素或为物体,或为背景*/
         outer=1;
         if(image[i-1][j]==0||image[i+1][j]==0||image[i][j-1]==0||image[i][j+1]==0)
              inner=0;
         if(image[i-1][j]==1||image[i+1][j]==1||image[i][j-1]==1||image[i][j+1]==1)
              outer=0;
         if(inner==0&&outer==0&&image[i][j]==1)/*像素周围既有物体又有背景*/               bianyuan[i][j]=1;/*,且该像素为物体上的像素(image[i][j]==1)，则定义为边界*/
      }
}

void output(int array[M][N],int n)
{
   int i,j;
   for(i=0;i   {
      printf("\n");
      for(j=0;j        if(array[i][j]==1)
           printf("1");
        else
           printf(" ");
   }
}

void main()
{
   int image[M][N]={{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                              {0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0},
                              {0,1,1,1,1,0,0,1,1,1,1,1,1,0,0,1,1,1,0},
                              {0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,0},
                              {0,0,1,1,1,1,0,0,0,1,1,1,1,1,0,1,1,1,0},
                              {0,1,1,1,1,1,1,0,0,1,1,1,0,0,1,1,1,1,0},
                              {0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0},
                              {0,0,0,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,0},
                              {0,0,1,1,1,1,1,1,1,0,0,0,0,1,1,1,1,1,0},
                              {0,0,0,0,1,1,1,1,0,0,0,0,0,1,1,1,1,1,0}};
   int bianyuan[M][N]={0};
   int i,j;
   printf("\nThe origianl image is:\n");
   output(image,10);
   edge(image,bianyuan);
   printf("\nIts edge is:\n");
   output(bianyuan,10);
}

写完了，又看一下，感觉edge函数太罗嗦了，不够简练，想了一下，改成了下面的样子，函数接口不变：

void edge(int image[M][N],int bianyuan[M][N])
{
   int i,j;
   for (i=0;i      for(j=0;j         bianyuan[i][j]=0;
   for(i=1;i      for(j=1;j      {
          int t=image[i-1][j]+image[i+1][j]+image[i][j-1]+image[i][j+1];
          if(t>0&&t<4&&image[i][j]==1)/*周围4个像素值介于1~3之间，*/
               bianyuan[i][j]=1;             /*且当前像素为物体，则其必为边界*/
      }
}

 

 

BugEyes 发表于 2005-11-18 16:52:00