点阵字库

little_cool_monkey: 原理没问题，B模块内call B模块回调函数，没有意义

cliffzhao: 你好，我想了一个算法，你看一下。思路是：不论哪条路径，横向的步数和纵向的步数之和都是固定的，所以可以把路径进行二进制编码为00011,01100,等等，其中0是向下走一步，1是向右走一步。 [code=cpp] void enum_path(int remain_length, int remain_width, int remain_height, ENUM_STEP_DIR step_dir, int prev_path){ if(step_dir == STEP_DIR_H){ prev_path |= 1; remain_width--; } else if(step_dir == STEP_DIR_V){ prev_path &= ~1; remain_height--; } remain_length--; if(remain_length == 0){ //a complete path has been enumerated g_v_path.push_back(prev_path); return; } prev_path = (prev_path 0) enum_path(remain_length, remain_width, remain_height, STEP_DIR_H, prev_path); if(remain_height > 0) enum_path(remain_length, remain_width, remain_height, STEP_DIR_V, prev_path); return; } [/code]

shanxifanshi: 写的不错

puppylpg: 赞啊！还是这篇文章详细实在清晰明了！

cigaiettes: 很好的学习例子.学习收藏了....