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该ppt包含了栈和队列的数据结构及基本操作，同时还有栈的举例(This ppt contains the stack and queue data structure and basic operation, as well as examples of the stack)

迷宫问题c源代码，大小为8*8的迷宫。主函数中输入开始位置和结束位置的坐标，输出所有可能的路径(Maze c source code, size 8* 8 of the maze. Enter the start the main function of the coordinates of the position and end position, the output of all possible paths)

字符串，实现了文件中查找想要的字符串，并将其标记出来

穷举算法，VC中的经典程序之一，适合初学者，(Brute-force algorithm, the the of the classic procedures in in the the VC, suitable for beginners)

数据结构课程设计，在VC环境下完成，利用哈夫曼树原理，主要功能是实现字符的压缩和解压缩，设计脉络清晰，易懂(Data structure course design, completed in the VC environment, using Huffman tree principle, the main function is to achieve the character of the compression and decompression, design, clear structure, easy to understand)

说明：  标准答案习题1 1.1 简述C++语言程序的结构特点。 答： （1）C++程序由一个或多个函数组成，其中至少有一个主函数main()，程序从主函数开始执行，由主函数来调用其他函数。(Standard answer to Exercise 11.1 on C++ Language program structural characteristics. Answer: (1) C++ Procedures by one or more functions, of which at least one of the main function main (), the procedure started from the main function, from the main function to call the other function.)

约瑟夫问题，用数据结构的循环链表实现约瑟夫问题的解答(Joseph problems，30 people in a circle, and inspires us with a circular chain to said. Can use structure array to make a cycle chain. The structure has two members, one for pointing down a pointer to a circular chain )

有向图的邻接表存储的实现，包括有向图的各种操作(Directed graph adjacency list storage implementation, including a variety of operations to the map)

A*算法 1、将开始节点放入开放列表(开始节点的F和G值都视为0) 2、重复以下步骤: 在开放列表中查找具有最小F值的节点,并把查找到的节点作为当前节点 把当前节点从开放列表删除, 加入到封闭列表. (A* algorithm 1, will begin to node placed in the and opening up list of (the began to node of the F and G values ​ ​ are regarded as 0) 2, repeat the the following steps: to Find a the has a the the smallest F value of the node in the the and opening up list of, and put Find a to the node as the current node current node is removed from the open list, added to the closed list.)

这是一个二叉树左右子树交换的简单程序，希望对大家有用！(This is a binary tree so the simple sub-tree exchange procedures in the hope for all of us!)