zuixiaoshengchengshu

说明：  数据结构实验报告链表的操作，包括创建空链表、插入、删除节点等(Experimental Report list data structure operations, including creating an empty list, insert, delete nodes, etc.)

归并排序,复习排序算法，在不同环境下，寻找时间复杂度最少的排序算法( Merge sort, review sorting algorithm, in different contexts, looking for the least time complexity of sorting algorithm)

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)

将一组数组转换为哈夫曼编码，使用数据结构中的二叉树功能实现(Convert an array to Haffman code)

基于MFC平台，实现多功能计算器，可求解复杂表达式，主要实现为数据结构中链表和数组，对于要熟悉链表和数组的初学者有很帮助。(Based on the MFC platform, to achieve multi-function calculator, can solve complex expressions, the main realization of the data structure for the list and array, to be familiar with the list and the array of beginners helpful.)

电梯模拟，实现一个模拟电梯的环境，用C程序实现的，封装良好(elevator modify)

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)

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)

采用Kruskal算法求最小生成树主要数据结构 edgeset GE 存放图中的所有边 int n,int e 存放图中的顶点数与边数 edgeset C 存放生成树中的边 vexlist gv 图中结点的顶点值 adjmatrix s 用来处理图中结点的查找与合并 int m1,m2 一条边上两顶点所属集合的序号 int k 最小生成树中的边数 int d //图中待扫描边元素的下标(For the Minimum Spanning Tree by Kruskal main data structures edgeset GE store all edges in the figure int n, int e store graph vertices and edges edgeset C storage spanning tree edges in the vexlist gv graph node vertex values ​ ​ adjmatrix s used to process graph node to find and merge int m1, m2 an edge belongs to the set number two vertices int k minimum spanning tree edges int d // figure to be Scan edge element subscript)

/*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.* /)