jgctgu

二叉树的C++实现，成员可以自己调整，实现了各种修剪此树的方法：初始化，增加节点，删去节点，复制树，读取节点，遍历树(Binary tree in C++, members can make their own adjustments to achieve a variety of pruning this tree methods: initialize, add nodes, delete nodes, copy the tree, read the node, traverse the tree)

说明：  标准答案习题1 1.1 简述C++语言程序的结构特点。 答： （1）C++程序由一个或多个函数组成，其中至少有一个主函数main()，程序从主函数开始执行，由主函数来调用其他函数。(Standard answer to Exercise 11.1 on C++ Language program structural characteristics. Answer: (1) C++ Procedures by one or more functions, of which at least one of the main function main (), the procedure started from the main function, from the main function to call the other function.)

克鲁斯卡尔算法思想.cpp int seekedge(EDGE * (&a), int n) void bubble(EDGE * (&a), int size) void searchandjoin(EDGE * (&a1), int k) // a1是图边集头指针，a2是最小生成树边集的头指针，k为所要删的边 void storepicture(EDGE * (&a), int n, VERTEX *b, int m) (Thinking of Kruskal algorithm. Cpp int seekedge (EDGE* (& a), int n) void bubble (EDGE* (& a), int size) void searchandjoin (EDGE* (& a1), int k) // a1 edge set is the first pointer map, a2 is the minimum spanning tree of the first indicators margination, k for the edges to be deleted void storepicture (EDGE* (& a), int n, VERTEX* b, int m) )

建立试题库文件，随机产生n个题目； 题目涉及加减乘除，带括弧的混合运算； 利用栈求解表达式的值； 随时可以退出； 保留历史分数，能回顾历史，给出与历史分数比较后的评价( The establishment of test database file, the subject of n randomly generated topics related to addition, subtraction, mixed operations with parentheses use the value of the expression stack solution ready to quit to preserve the historical scores, can look back at history, and history gives the comparison of scores Evaluation 保留历史分数，能回顾历史，给出与历史分数比较后的评价)

.h文件。c++中的复数类，能够直接完成基本四则运算，输入输出流，完成一定范围内的比较运算(. H file. c++ classes in the plural, to directly perform basic arithmetic, input and output streams, to complete a range of comparison operators)

一道poj上的问题的解答，棋盘上的距离问题。即实现棋盘上从一个点移动到王，后，兵之间最短步数的解答。(A poj answers to your questions on the chessboard distance problem. That is, move from one point to the king on the board, after the shortest number of steps between the soldiers answer.)

各种排序的模板实现，并分析了各种实现的时间复杂度及改进办法(Templates of various sort, and analyze the time complexity of the various implementations and ways to improve)

普利姆算法用来求解最小生成树问题很容易，使得树的最小权值最小。(Prim algorithm used to solve the minimum spanning tree problem is easy, the minimum weight is the lowest of the tree)

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.)

用邻接表表示的数据结构，及DIJKSTRA算法实现的最短路径程序(adjoining table with the data structure and algorithm DIJKSTRA the shortest path procedures)