困扰了好几天的一个算法
求一个算法 思路
如上图，大概是一个 “against”字样，由“1”组成，然后再它的周边还有一些零散的“1”，我现在想把这些零散的“1”去掉。我的思路是想通过计算连接在一起的“1”的数量，当它的数量未超过一个常数时，将它们置为0，以取到存的字块。
简单说，我就是想 获取由“1”组成的图案有多少个“1”，比如图中的a由多少个“1”组成 。这个算法要怎么写，本人算法是在是太烂了
回复讨论(解决方案)
高斯降噪算法
以每个点的周围，看看是否是连续走三步的（都可以有个1可以接班的）。不是，就要删除。
以每个点的周围，看看是否是连续走三步的（都可以有个1可以接班的）。不是，就要删除。
开始是用递归连续检查，结果发现效率太低了 ...
高斯降噪算法
百度，谷歌上的资料好少呀，能提供一个实例代码的链接吗？
我觉得只能用图形相关算法，数学算法，无法避免把 against 中小写 i 上面的点误杀
我觉得只能用图形相关算法，数学算法，无法避免把 against 中小写 i 上面的点误杀
有什么好的推荐吗?
木有，对图形不熟
可以去 图形图像/机器视觉 http://bbs.csdn.net/forums/ST_Image 问
我想到的只是用字体生成against图，然后和原图做相差，但貌似也不是个好的解决方法
木有，对图形不熟
可以去 图形图像/机器视觉 http://bbs.csdn.net/forums/ST_Image 问
我想到的只是用字体生成against图，然后和原图做相差，但貌似也不是个好的解决方法
哦哦，anyway，thx
你的需求可能与这个有关 http://bbs.csdn.net/topics/390438979
这个我搞过，和你思路一模一样，我是当成数组，然后把每个字符逐行逐列扫描的，然后统计百分比。其实效率低的吓人
你的需求可能与这个有关 http://bbs.csdn.net/topics/390438979
膜拜！
你的需求可能与这个有关 http://bbs.csdn.net/topics/390438979
膜拜！
原理大概都是一样的，软件里面所说的学习，无非是增加验证码库，我觉得主要困难的还是在前面的去除干扰，提取出比较纯的字块上
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