qiongju

这是birch C语言的源代码。里面有详细的代码注释方便学习birch算法(birch c)

用于迷宫问题的解决用户根据根据要求进行设置即可在屏幕上显示(Maze problems for users in accordance with the requirements set can be displayed on the screen)

① 实现在B-树上的查找，并分析其时间复杂性。 ② 实现B-树的ADT，包括其上的基本操作：结点的加入和删除。 ③ 要求B-树结构中的M=3 或5，实现其中的一种即可。 ④ 实现基本操作的演示。 (4) 实现提示 (b-trees c/c++)

最小生成树C语言程序，提供给喜欢编程以及数学建模的同学(Minimum spanning tree C language program)

蜂群算法C语言实现，附带中文注释，简单易懂(ABC algorithm coded using C programming language)

经典数据结构作业，黑杰克游戏。要求程序可以连续打10次，每场比赛的赢家，赢得一个点（两点一倍）。在结束的10场比赛为你和电脑打印出的总奖金。(Write a program that simulates a simplified version of the game "BlackJack". A deck of cards numbered 2 to 10, jacks, queens, kings, and aces is used, where jacks, queens and kings are counted as 10s, and aces are either 1 or 11 at your choice. )

这是acm高手楼天城的比赛代码，很有用哦，对于提高算法非常有用(This is a master of building Tiancheng acm game code, very useful oh, for improving the algorithm)

这个是关于数据结构中冒泡排序算法的实现，大家可以利用这个为基础，进一步研究和利用排序(This is a data structure on the Bubble Sort algorithm, we can use this as a basis for further study and use of sorting)

/*快速排序：以待排序元素中的第一个元素作为基准将带排序的元素分为三组 *大于基准元素为一组置于基准值的右边，小于基准值的元素为一组置于基准值的左边， *基准值为一组，这样就找到了基准元素的位置就确定了 *如此递归调用。直到每一组的low==high，即实现了排序的要求。 */ (/* Rapid Sort: the first element to be sorted in the element as a basis with with sorting the elements are divided into three groups* is greater than the reference element of a group is placed on the right side of the reference value, smaller than the reference value elements for a group of placed on the left side of the reference value,* reference value, a group, so that the location of the reference element is found it is determined* so recursive call. Until each group of low == high to achieve the requirements of the sort.* /)

迷宫求解一般采用“穷举法”，逐一沿顺时针方向查找相邻块（一共四块－东(右)、南(下)，西(左)、北(上)）是否可通，即该相邻块既是通道块，且不在当前路径上。用一个栈来记录已走过的路径栈是限定仅在表尾(top)进行插入或删除操作的线性表。(Maze solving generally use the " exhaustive" one by one to find adjacent blocks in a clockwise direction (a total of four- East (right), South (down), West (left), North (on)) whether to pass, that the both channel blocks adjacent blocks, and not in the current path. Use a stack to record the path traversed stack is limited only in the tail (top) to insert or delete a linear form.)