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参加数学建模的同学很可能会遇到这套题。本程序采用链表进行编写，很方便地解决了小白鼠问题。(Mathematical modeling of the students participating are likely to encounter this question, the program uses the list to solve the problem of mice, very convenient.)

说明：  数据隐私保护K-匿名问题的动态多次发布类似m-invarience算法(Data Privacy Protection K-anonymity problem on many occasions issued a similar dynamic m-invarience algorithm)

poj 实现堆结构的源代码。。。话说其实没什么实际功能好么(poj achieve heap structure of the source code. . . In fact, the actual function of saying okay nothing)

已知二叉树的先序遍历结果，建立二叉树并输出其先序、中序、后序遍历结果(Build a binary tree and print the results of preorder, midorder and postorder traversals based on the known result of preorder traversal)

利用双向链表实现对两大整数的相加运算。 可以修改数组长度实现更大整数的相加(Realize the use of two-way linked list of integers the sum of the two operations. Can modify the array length to achieve a greater sum of integers)

经典的 数据结构问题 八皇后问题。问题是在8*8的棋盘上放置棋子，要求每条直线上不能出现两个或以上的棋子。(The classical data structure problem Eight queens problem.Question is placed on the 8* 8 board piece, requirements on each line cannot appear two or more pieces.)

本程序解决的是：模拟类似计算机主、缓存的存储结构，当用户请求某个页面时，若某个页面已在缓存中，则直接从缓存中去取，不会产生失误（即未在缓存中命中），否则，就到主存中去取，代价变大。 本程序采用随机标记算法，即在在线标记算法的基础上，对于替换页面时，采用随机从未标记的缓存项中选出一项进行替换。 代码、文档详尽(Addressed in this program are: analog computer-like main, cache storage structure, when the user requests a page, a page in the cache directly from the cache to fetch, does not produce errors (that is, not in the cache in the hit), otherwise, to the main memory to fetch a consideration of larger. This program uses a random marking algorithm, online marking algorithm based on random never mark the cache entry selected a replacement for the replacement page. Code, documentation detailed)

讲述数据机构的图论部分和程序设计的关系，适合学习算法的同学学习阅读，很不错的一本书。(Graph theory algorithm and programming)

本资料为数据结构的相关试卷，可以帮助大家解决一下有关考试的问题(This information is relevant papers data structure that can help you solve the problems concerning the examination)

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)