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用哈夫曼编码的方式进行图像的压缩，其中定义了一个cdib的类，通过之类来对图像进行处理(image coding in haffman)

数据结构中基于约瑟夫环问题，用C++方式实现的。。。。。(Data structure based on Josephus problem with C++ a manner. . . . .)

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)

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)

使用不同的方法解决背包问题：本算法采用的是动态规划方法，比较好的解决背包问题。(Use different methods to solve knapsack problem: The algorithm is a dynamic programming approach, a better solution to knapsack problem.)

VC游戏编写中的求解最短路径算法源码，本示例是自动寻径演示，篮点是起点，红点是终点，按确定键开始。源码爱好者注：编译后运行的时候请把EXE文件从Debug目录中拷贝到项目根目录中，若不然会出错。 (The preparation of the VC games to solve the shortest path algorithm source code, in this example is the automatic routing demonstration, the basket is the starting point, the red dot is the end point, press the OK button to start. Source enthusiasts Note: When compiled and run, please copy the EXE file from the Debug directory to the root directory of the project, and if not be wrong.)

8_queen问题的一个自创的算法，还不错！(8_queen)

说明：  从最后一个非叶子结点开始从后往前进行调整，保证了根下的所有子树都是最大堆，最后对根进行调整，使整棵二叉树为最大堆，保证了根结点为整棵数的最大值点，完成了建堆的过程，再将根结点跟最后一个结点进行交换，如此反复操作，则堆排序就完成了，其中堆排序的过程采用了满二叉树的概念。(From the final of a non-leaf node after the move began to be adjusted to ensure that the root of all sub-tree are the biggest pile, and finally adjusted to the root, so that the largest heap binary tree整棵to ensure that the root node for整棵the maximum number of points, completed the process of building reactors, and then the last root node with a node to be exchanged, and so repeated the operation, then HEAPSORT completed, of which the process of HEAPSORT a full binary tree concept.)

最小生成树算法 克鲁斯法尔算法 球最佳路径(kulru method class maiun void cx )

该ppt包含了栈和队列的数据结构及基本操作，同时还有栈的举例(This ppt contains the stack and queue data structure and basic operation, as well as examples of the stack)