插入排序

直接插入排序算法验证。 快速排序算法验证。 直接选择排序算法验证。 (Direct insertion sort algorithm validation. Quick sort algorithm validation. Direct selection sort algorithm validation.)

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)

一个经典的旅行商问题源码，不是最优解，只是一种 思路(A classic traveling salesman problem source is not the optimal solution, just a thought)

数据结构试验-打印输出计算机本科专业4年每学期的课表；C++实现；图的邻接表存储结构以及拓扑排序的基本思想(Data structure test- printout computer 4 years of undergraduate per semester timetables C++ achieve basic idea of ​ ​ the adjacency table storage structure and topological sorting)

对于数据结构每一章节的相应的练习题，巩固好数据结构(To consolidate data structure corresponding practice questions for each chapter of the data structure)

实现栈的基本操作，建栈，进栈，出栈，数据结构实现(To achieve the basic operation of the stack, built stack into the stack, the stack data structure)

1>实现可以随意浏览任一景点的介绍。（功能一） 2>可以给出任意两个景点的最短距离及其路线。（功能二） 3>可以查看任一景点热度，并可以对对所有景点热度（访问次数）进行排序。（功能三） 4>每个景点相关内容从文件读取，实现对文件操作的练习。 (1> implementations can feel free to browse introduce 任一景 points. (Function a) 2> You can give the shortest distance and route between any two attractions. (Function b) 3> You can view Renyi Jing point heat, and heat can all attractions (visits) to be sorted. (Function c) 4> each spot related content file reading, to achieve the file operations exercises.)

说明：  链表逆转,实现单链表的逆转程序，数据结构(List Inverse)

大一c语言实验的所有源码以及实验报告报告(All source code and report experimental freshman c language)

约瑟夫环（约瑟夫问题）是一个数学的应用问题：已知n个人（以编号1，2，3...n分别表示）围坐在一张圆桌周围。从编号为k的人开始报数，数到m的那个人出列；他的下一个人又从1开始报数，数到m的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。通常解决这类问题时我们把编号从0~n-1，最后结果+1即为原问题的解。 (Josephus (Josephus problem) is the application of a mathematical problem: Given n individuals (with numbers 1,2,3 ... n respectively) sitting around a round table. From the number of people gettin k, number of the m man out of the line he s the next person and a number of gettin number to m the man was out of the line and so the law is repeated until the round table were all out of the column. We numbered 0 ~ n-1, the final result is the original problem solution+1 usually solve these problems.)