这个是数据结构中一定有的哈夫曼的实验

一个简单的公交查询管理系统，采用最短路算法，利用odbcapi函数访问数据库。-a simple inquiry transit management system, the most shorted algorithm, use odbcapi function access to the database.

利用C++对MYSQL进行数据读，写，及修改. 需要有mysql.h 头文件   数据库结名为person, 表名为 mm, 表结构如下：  字段             类型 device_id       int sav_value      int sav_date        datetime

BFS、DFS、有向图、无向图中的各种算法的实现，可以自动生成图形。-BFS, DFS, to the map, no map to the realization of the algorithm can be automatically generated graphics.

将两个随机产生的链表连接,并按从小到大的顺序输出-two randomly generated Chain Link, together with a small to large export order

一个用VC开发得公交车线路查询系统！-a VC development in the bus line query system!

利用Fortran 90 接口读取已经创建的nc文件，并且检验其读写是否正确-The use of Fortran 90 interface to read nc file has been created, and test its read and write correctly

抽奖小玩意,我的简单vf应用!-plaything of the draw, I use a simple vf!

1.[问题描述] 编写递归算法，计算二叉树中叶子结点的数目 [输入] 按照先序序列的顺序输入该结点的内容。其输入abd eh cf i g . [输出] 按中序序列输出，输出的结果为；dbheaficg并计算出二叉树中叶子结点的数目为4 [存储结构] 采用二叉表存储 [算法的基本思想] 采用递归方法建立和遍历二叉树。首先建立二叉树的根结点，然后建立其左右子树，直到空子树为止，中序遍历二叉树时，先遍厉左子树，后遍厉右子树，最后访问根结点。根据左右子树的最后一个结点计算出二叉树中叶子结点的数目。 程序如下： #include #include #include"stdlib.h" -1. [Description of the issue] to prepare recursive algorithm, Binary calculation leaves the number of nodes [imported] in accordance with the first order input sequence in the order of the node contents. Input abd eh i g cf. [Output] by the order sequence output, the results of the output; dbheaficg calculated Binary leaf node to the number four [storage structure] Table 2 forks storage [ The basic idea algorithm] recursive method and traverse binary tree. First established binary tree root node, and then to build their son around the tree, the tree until the loopholes, which preorder binary tree, Li first times the left sub-tree, right af

迷宫求解问题。程序中已经建好了一个数组作为迷宫数组。然后进行试探，输出迷宫从起点到终点的路径。-Maze solving the problem. The program has been built as a maze of an array of arrays. And then test the output from the beginning to the end of the maze path.

sybase数据库同步备份1采用了tcp通信。 2实现自由选择主辅库。