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说明：  100米比赛排名 C++实现100米比赛排名 C++实现(100m rankings C++, the 100 meters ranks C++,)

输入迷宫行列数，1代表墙0代表路，之后系统生成路径(Enter the maze ranks number 1 represents 0 for Wall Road, after the system-generated path)

数据结构与算法分析的 C++检索代码，很经典(the code of Data structure)

对于给定的n个字符串，请找出一种最佳的连接方式，使得采用该连接方式得到的字符串在所有连接的结果中，字典序最小(For a given n-string, find an optimal connection, making use of the connection string to get the results in all connections, the lexicographic minimum)

C语言树的简单实现,介绍了树你前序遍历，树插入与删除节点的实现(C-tree is simple to achieve, before you introduced the tree traversal, tree insert and delete nodes to achieve)

数据结构中图的存储结构的建立与搜索 建立图的二种存储结构 在不同的结构上实现遍历 (Data structure diagram of the storage structure of the establishment of the search to establish two kinds of graph storage structure in the different structures to achieve through)

选择排序的实现，是一种基本的排序方法，但是时间复杂度相对较大。(Select Sort realization, is sort of a basic, but relatively large time complexity.)

使用C语言编写的AOV网的拓扑排序，内附有详细的源码解释和输入示意图(Using C language AOV network topological sorting, enclosing a detailed explanation and input source diagram)

用模板实现了单向链表的增删改查等功能，并且适用于多种数据结构和类，简单实用，是学习和使用的绝好资料(Template to achieve a one-way linked list CRUD functions, and is suitable for a variety of data structures and classes, simple and practical, is to learn and use excellent information)

说明：  拓扑排序 　 对一个 有向无环图 G进行拓扑排序，是将G中所有顶点排成一个线性序列，使得图中任 意一对顶点u和v，若u，v ∈E，则u在线性序列中出现在v之前。 　 通常，这样的线性序列称为满足拓扑次序的序列，简称 拓扑序列 。 注意： 　 ①若将图中顶点按拓扑次序排成一行，则图中所有的有向边均是从左指向右的。 　 ②若图中存在有向环，则不可能使顶点满足拓扑次序。 　 ③一个DAG的拓扑序列通常表示某种方案切实可行。 (Topological sort In a directed acyclic graph G topological sort, is all vertices will G formed a linear sequence, make diagram post Meaning a pair of vertex u and v, if u, v ∈ E, then u in linear sequence in which they appear in v before. Usually, such linear sequence sequence of topological order called meet, abbreviation topology sequence. Note: (1) if the diagram according to the topology order line vertices, then figure all from left to right edge are pointing. (2) if the figure to exist in the ring, is impossible to have the vertex meet topological sequence. (3) a DAG topological sequence usually expressed some scheme is feasible. )