StackandQueue

数据结构与程序设计中的重点例子程序：八皇后问题(Data structure and procedures of the focus of the design example of the procedure: eight queen problem)

说明：  一个用VS2008编写的冒泡排序算法演示程序，数组都是随机生成的。(Bubble Sort)

根据队列搜索算法的基本思想，现设计了一个简单的结构体数组如下定义代码所示。此结构体数组具备控制结点搜索和存储结点信息和路径的功能，可以较好的实现TDN的最小时间路径搜索。(Search algorithm based on the basic idea of ​ ​ the queue is now designed a simple structure array as defined code. This structure array with the control node and storage node information search and path functions, you can achieve better TDN minimum time path search.)

PRIM算法 对任意给定的网和起点，用PRIM算法的基本思想求解出所有的最小生成树。(PRIM algorithm for any given network and the starting point, PRIM algorithm used to solve the basic idea of all the minimum spanning tree.)

参加运动会的 个学校编号为 。比赛分成 个男子项目和 个女子项目，项目编号分别为 和 。由于各项目参加人数差别较大，有些项目取前五名，得分顺序为7，5，3，2，1；还有些项目只取前三名，得分顺序为5，3，2。写一个统计程序产生各种成绩单和得分报表。 2、要求产生各学校的成绩单，内容包括各校所取得的每项成绩的项目号、名次（成绩）、姓名和得分；产生团体总分报表，内容包括校号、男子团体总分、女子团体总分和团体总分。 3、测试数据：对于 ， ， ，编号为奇数的项目取前五名，编号为偶数的项目取前三名，设计一组实例数据。 (The school serial number participating in Games is. Competition divides into men s event and women s event , the project serial number parts for the sum. The difference is bigger since every project participates in a number , some projects choose the first five , score order is 7 , 5 , 3 , 2 , 1 Still have some of projects taking the first three places , score only being 5 , 3 , 2 in proper order. Write a form for report counting procedure producing the various school report card and score. 2, demands the school report card producing every school , content includes every achievement project number , position in a name list (achievement) , full name and score per got by school Produce the group total points form for report , content including the school number , male person group total points , woman group total points and group total points. 3, testing data: Be that the odd number project chooses the first five to the serial number, the serial number is that the even number project c)

用链表解决约瑟夫问题。 约瑟夫环是一个数学的应用问题：已知n个人（以编号1，2，3...n分别表示）围坐在一张圆桌周围。从编号为k的人开始报数，数到m的那个人出列；他的下一个人又从1开始报数，数到m的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。 其中包括一个实验报告，介绍了编程思路和输出结果截图。(List to resolve Joseph. Josephus is the application of a mathematical problem: Given n individuals (numbered 1, 2, 3, ... n, respectively) sitting around a round table around. From number k people began to count off the number to m the man out of the line his next person and from a number off, the man was out of the line number to m so regularly repeated down until roundtable around of the people all of the columns. Including a lab report describes the programming ideas and output screenshots.)

简单的汉诺塔算法，采用归约算法的思想，实例是三个盘子，三个柱子(Simple HANOR algorithm)

根据队列搜索算法的基本思想，现设计了一个简单的结构体数组如下定义代码所示。此结构体数组具备控制结点搜索和存储结点信息和路径的功能，可以较好的实现TDN的最小时间路径搜索。(Search algorithm based on the basic idea of ​ ​ the queue is now designed a simple structure array as defined code. This structure array with the control node and storage node information search and path functions, you can achieve better TDN minimum time path search.)

单向链表的创建与操作 设单向链表中节点的数据域的数据类型为整型，编写函数实现以下操作： （1）实现单向链表的创建（包括初始化）与输出操作，节点的个数及节点的数据由用户输入。 （源代码：ds3-1.c） （2）查找给定的单链表中的第i个节点，并将其地址返回。若不存在第i个节点，则返回空地址。 （源代码：ds3-2.c） （3）查找给定的单链表中值为n的节点，并将其地址返回。若不存在值为n的节点，则返回空地址。同时，还应通过参数传回该节点的序号。 （源代码：ds3-3.c） （4）删除给定的单链表中的第i个节点，成功返回1，失败返回0。 （源代码：ds3-4.c） （5）删除给定的单链表中值为n的节点，成功返回1，失败返回0。 （源代码：ds3-5.c） （6）在给定的单链表的第i位上插入值为n的节点。 （源代码：ds3-6.c） （7）在给定单链表的值为m的节点的前面插入一个值为n的节点。 （源代码：ds3-7.c） (Creation and operation of a one-way linked list Set up a one-way linked list data type node integer data fields , write a function to achieve the following: ( 1 ) achieve the creation of a one-way linked list ( including initialization ) and output operation , the number of nodes and node data entered by the user . ( Source : ds3-1.c) ( 2 ) Find a single list given in the i-th node and returns its address . Without the presence of the i-th node , returns an empty address. ( Source : ds3-2.c) ( 3 ) Find a given node in a given value of n single list , and return address . Without the presence of the value of n nodes , returns an empty address. Meanwhile, the number should be returned by the parameters of the node . ( Source : ds3-3.c) ( 4 ) Delete the given singly linked list in the i-th node , the successful return 1, else return 0 . ( Source : ds3-4.c) ( 5 ) to delete a single node in the list is given n , the successful return 1, else return 0 . ( Source : ds3-5.c) ( 6 ) )

一个很经典的问题_八数码! 算法速度快,解决问题多...有兴趣的可以(A very classic problem _ eight digital! Fast algorithm to solve the problem ... are interested in more than can be)