实现了二叉排序树的所有功能 建立，增，删，改，查

论坛中有许多操作Access数据库的介绍、说明，示例代码等资源，但感觉总是不太连贯和全面，给急需想要使用Access数据库做开发的朋友不少遗憾。本代码是一个封装完整的Access数据库类，包括Ole版本协商、数据库优化、数据库连接等等。由于接口非常简单易懂，读者下载后即可直接投入使用，非常方便。不过，由于Access DAO不支持多线程，所以在本代码中没有实现更好的替代方案，这也是读者在使用本代码的过程中需要注意的一个地方。由于本人水平有限，程序中难免会有很多缺陷，还请各位高手指点！

应用背景图论相关，图论相关算及其应用。邻接表，邻接矩阵 的实现方法，。。图论相关，图论相关算及其应用。邻接表，邻接矩阵 的实现方法图论相关，图论相关算及其应用。邻接表，邻接矩阵 的实现方法关键技术邻接表，邻接矩阵 的实现方法，C++源码 源代码，使用 VS 2010 C++源码 源代码C++源码 源代码 编写的，真麻烦

一些简单的小程序，如哈弗曼编辑器，最小生成树，哈希表的设计与实现-Some simple small programs, such as Havermann editor, minimum spanning tree, hash table design and implementation, etc.

本目录存放的小波编码实现的源程序： 打开ezw2.dsw工作区，编译后可在ezw2ezwdebug中找到编码可执行程序； 在ezw2unezwdebug中找到解码可执行程序。-Wavelet realization of the source coding : Open ezw2.dsw work, compiled in ezw2 ezw debug find executable code; In ezw2 unezw debug find decoder executable.

KRUSKAL算法-KRUSKAL Algorithm

此函数的功能是依次选出两个小的数，不要求原来数组中的数按大小顺序排列。-The function of this function are followed by two smaller number of elected and does not require a number of the original array by the size of order.

自己学VB时数据库编程的一个小程序-studied VB Database Programming a small program

icarnegie 课程的sdd7的练习2，满分，决对正确，放心使用-icarnegie practice sdd7 courses 2, perfect, the right decision, rest assured that the use of

VC编写的读取数据文件的源码-VC compilation reads the fetching according to the document source code

迷宫算法（maze） /* Maze * Starting point is m[0][0], need to find a path go to m[9][9]. 0 means OK, * 1 means cannot go there, boundary is 0 and 9, cannot go beyond boundary. * Each step can be made horizontally or vertically for one more grid (diagonal * jump is not allowed). * Your program should print a series of grid coordinates that start from m[0][0] * and go to m[9][9] * Hint: No need to find the shortest path, only need to find one path that gets * you to desitination. */ -/* Maze * Starting point is m[0][0], need to find a path go to m[9][9]. 0 means OK, * 1 means cannot go there, boundary is 0 and 9, cannot go beyond boundary. * Each step can be made horizontally or vertically for one more grid (diagonal * jump is not allowed). * Your program should print a series of grid coordinates that start from m[0][0] * and go to m[9][9] * Hint: No need to find the shortest path, only need to find one path that gets * you to desitination.