hellow4

简单的一个链栈.即用链表实现的一个栈。可以做简单的插入、删除、查找等操作(this is a simple C++ program )

描述：足球俱乐部包括球员、普通行政人员。球员有姓名、年龄、能力、年薪、转会费、进球总数、服役年限等重要参数；普通行政人员有姓名、年龄、能力、年薪等参数。你需要给相关管理机构开发一个管理程序，实现对众多足球俱乐部管理的基本功能。具体功能包括： a) 俱乐部的增删改查，俱乐部的属性包括名称、现金、其下人员； b) 可以对所有球员的各项属性进行简单搜索，支持输入多个条件，条件之间用&&连接，例如： i. “年龄不大于25且服役年限大于5年且进球数大于100的球员”，查询表达式：!(@age>25) && @experience>5 && @kick>100 c) 支持球员在不同俱乐部间的买卖，买卖需符合以下条件：作为买方的俱乐部的现金>=球员的转会费，交易完成后，卖方将得到转会费。 要求： d) 必须采用面向对象的方式，使用类组织数据结构，并注意类的继承关系； e) 在题目要求基础上可以自行扩展功能，提供扩实用扩展功能者适当加分（加分不超过5分）。 (Description: Football club including players, general administrative staff. Player name, age, ability, salary, transfer fee, total number of goals, service life and other important parameters ordinary administrative staff name, age, ability, salary and other parameters. You need to give the relevant management agencies to develop a management program to achieve the number of football club management s basic functions. Specific features include: a) the club CRUD, the name of the club s property, including cash, under which personnel b) all players can perform a simple search of the property, supports input multiple conditions with && connection between conditions, for example: i. "Age is not greater than 25 and greater than 5 years service life and goals more than 100 players," query expression:! (@ age> 25) && @ experience> 5 && @ kick> 100 c) Support the players traded between different clubs, the sale subject to the following conditions: As a buyer s club cash> = pl)

在VC中实现Bresenham算法画直线(Bresenham algorithm in VC draw a straight line)

随机函数产生10000个随机数，用快速排序、直接插入排序、冒泡排序、选择排序的排序方法排序，并统计每一种排序所花费的排序时间和交换次数。其中随机数的个数由用户定义，系统产生随机数。(Random function generates 10,000 random numbers using quick sort, insertion sort, bubble sort, select Sort sort Sort, and statistics for each sort spend the sort of time and the number of exchanges. Wherein the number of the random number defined by the user, the system generates random numbers.)

红黑树 二叉树 性能比较分析 红黑树 二叉树(Red-black tree binary tree performance analysis,)

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.)

可以实现判断二叉树是否为完全二叉树、判断两棵二叉树是否镜像对称、求一棵二叉树的一个最长枝等功能(Can determine whether the binary tree is complete binary tree to determine whether two binary mirror symmetry, one of the longest seeking a binary tree branches and other functions)

图的邻接表生成，深度遍历及任意两点间的路径(The adjacency table to generate deep traversal path between any two points)

说明：  简单的数据结构算法 在POJ上的题目 利用C++实现(Simple data structure algorithm in POJ on the subject using C++ Implementation)

自己做的一个用用栈实现类的操作，是课内作业，希望和大家互相交流(Own a type used by the realization of the operation of the stack is class work, hope and we can exchange ideas)