lianbiao

马踏棋盘的贪心算法，8*8棋盘，基本上不用回溯。(horse riding chessboard of the greedy algorithm, 8* 8 chessboard, basically not retroactive.)

最小生成树两种经典算法,kruscal和Prim算法(spanning tree)

Monkeys and peach,Monkeys and peach(Monkeys and peach)

该程序是数据结构中以链表存储形式来求两个集合的交集，用C++语言描述的(The program is the form of a linked list stored in the data structures to seek the intersection of two sets, C++ language to describe)

数据结构，关键路径问题，设计并实现关键路径的一种应用，采用图的算法(Data structure, critical path problem, using graph algorithm)

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.)

数据结构 堆栈 实现了以下功能： cout<<"----------------主菜单---------------"<<endl <<"* 1.建立 *"<<endl <<"* 2.输出 *"<<endl <<"* 3.判断栈是否为空 *"<<endl <<"* 4.栈的长度 *"<<endl <<"* 5.输出栈顶元素 *"<<endl <<"* 6.进栈 *"<<endl <<"* 7.出栈 *"<<endl <<"* 8.遍历 *"<<endl <<"* 9.数制转换 *"<<endl <<"* 10.清空 *"<<endl <<"* 11.销毁 *"<<endl <<"* 12.退出 *"<<endl <<"--------------------------------------"<<endl<<endl (data structure stack)

说明：  二叉树的演示程序。实现了二叉树和哈弗曼树(Binary tree demo program. Implements a binary tree and the tree Hafu Man)

大二时做的一个数据结构的课程设计，用2-3树实现的图书馆管理系统，只是控制台程序，没有很好的图形界面。包括管理员、读者。管理员功能有：采编入库、查看图书、查看读者、删除读者。读者功能有：借书、还书、预约、查找等等。。功能全面。包括注册账号，登录需要账号密码。(Sophomore to do a data structure of the curriculum design, with 2-3 tree implementation of library management system, but the console program, not a good graphical interface. Including administrators, readers. Administrator functions include: editing storage, check books, check readers, delete readers. Reader functions: library, book, booking, search and so on. . Full-featured. Including the registration account, login account password required. )

在N*N的棋盘上(1<=N<=10)填入1,2,...N*N共N*N个数,使得任意两个相邻的数之和为素数. (In the N* N chessboard (1 < = N < = 10) filled 1,2, ... N* N total number of N* N so that any two adjacent numbers and prime numbers.)