用无向网表示学校的校园景点平面图，图中顶点表示主要景点，存放景点的编号、名
称、简介等信息，图中的边表示景点间的道路，存放路径长度等信息。要求能够回答有关景
点介绍、游览路径等问题。游客通过终端可询问：
（1）从某一景点到另一景点的最短路径。
（2）游客从公园进入，选取一条最佳路线。
（3）使游客可以不重复地浏览各景点，最后回到出口（出口就在入口旁边）。
[基本要求
The undirected network is used to represent the campus scenic spot plan of the school. The vertex of the map represents the main scenic spot, and the number and name of the scenic spot are stored.
The edges in the graph represent the roads between scenic spots and store information such as path length. Ask to be able to answer the scene.
Points introduction, tour routes and other issues. Tourists can inquire through the terminal:
(1) The shortest path from one scenic spot to another.
(2) Tourists enter the park and choose the best route.
(3) To enable visitors to visit the scenic spots without repeating them, and finally return to the exit (the exit is next to the entrance).
[Basic requirements (2019-07-06, C/C++, 812KB, 下载0次)
问题描述
若用无向图表示高速公路网，其中顶点表示城市，边表示城市之间的高速公路。试设计一个找路程序，获取两个城市之间的所有简单路径。
基本要求
（1） 输入参数：结点总数，结点的城市编号（4位长的数字，例如电话区号，长沙是0731），连接城市的高速公路（用高速公路连接的两个城市编号标记）。
（2） 输入 要求取所有简单路径的两个城市编号。
（3） 将所有路径（有城市编号组成）输出到用户指定的文件中。
Problem descriptionIf use undirected graph said freeways, with vertex said city, edge said highway between the two cities.Try to design a route program, obtain all simple paths between the two cities. The basic requirements(1) input parameters: the total number of nodes, the nodes of the city number (4 digits long Numbers, telephone area code, for example, changsha is 0731), connected to the city expressway (with high-speed highway connects the two cities of number tags).(2) input requirements take number all simple paths of the two cities.(3) will all paths (a number of city) output to the user specified file. (2017-01-02, Dev C++, 3KB, 下载4次)
管道铺设施工的最佳方案问题（需要在某个城市n个居民小区之间铺设煤气管道，则在这n个居民小区之间只需要铺设n-1条管道即可。假设任意两个小区之间都可以铺设管道，但由于地理环境不同，所需要的费用也不尽相同。选择最优的方案能使总投资尽可能小，这个问题即为求无向网的最小生成树。）本程序即是解决上述问题的，本人自己所写。
The best solution to pipeline construction issues (need a residential area of the city between the n laying gas pipelines, in the residential area between the n n-1 requires only laying the pipeline can be. Assumption of any two cells Rooms are laying pipes, but due to the geographical environment is different, the required fees are not the same. Choose the best solution to make a total investment of as little as possible, the problem is the undirected minimum spanning tree network.) This program that is, to solve these problems, I have written yourself. (2013-11-09, Visual C++, 3KB, 下载7次)
这是上海铁路局目前仍在使用的行包托运软件中的一部分内部算法。该题目采用1995年年底我国铁路运输网的真实数据进行编程和运行验证。
铁路运输网络中由铁路线和火车站的两个主要概念，譬如：1号铁路线表示京广线，2号铁路线表示京沪线等。
铁路线对象包括铁路线编号,铁路线名称，起始站编号,终点站编号,该铁路线长度，通行标志(00B客货运禁行,01B货运通行专线，10B客运通行专线，11B客货运通行)。
火车站对象包括所属铁路线编号，车站代码，车站名，车站简称，离该铁路线起点站路程及终点站路程。
2.基本功能
（1）查询某站所属的铁路线 （2）要求具备新增铁路线的管理功能
（3）要求具备新增车站的管理功能 （4）针对客运，货运情况能计算任何一个起始车站到任何一个终点站之间的最短路径。并且要求能够显示出该最短路径的各个火车站的经由顺序
3.输入输出
初始数据是从views.txt、lines.txt、ways.txt三个文件中读入，在读入数据后，用户可以根据选项选择相应的功能，不同的功能有不同的数据输入/输出，比如：查询的功能是要求输入要查询的站的名称，然后输出是该站的相关信息；查询最短路径的功能则是输入起点站、终点站的名称，，输出则是该的路线距离和经由站等
This is part of the internal algorithm in the Shanghai Railway Bureau is still in use in the line package consignment software. The topic using real data of the end of 1995, China s railway transport network programming and runtime verification.
The two main concepts of the railway line and railway station by rail transport network, such as: the railway line on the 1st Beijing-Guangzhou line, on the 2nd railway line, said the Beijing-Shanghai line.
Railway lines, including the railway line number, the name of the railway line, the starting point number, terminal number, the length of the railway line, traffic signs (00B passenger and freight forbidding, freight traffic Line 01B, passenger traffic Line 10B, 11B passenger and freight traffic) .
The railway station, including their respective railway line number, station code, station name, station referred to as the distance from the starting station of the railway line and the terminal away.
(2) the basic functions
(1) to query a s (2012-06-21, Visual C++, 461KB, 下载6次)
校园导游咨询
（1）基本要求：
① 设计你的学校的校园平面图，所含景点不少于10个。以图中顶点表示学校各景点，存放景点名称、代号、简介等信息；以边表示路径，存放路径长度等相关信息。
② 为来访客人提供图中任意景点的问路查询，即查询任意两个景点之间的一条最短的简单路径。
③ 为来访客人提供图中任意景点相关信息的查询。
（2）测试数据：由读者根据实际情况指定。
（3）实现提示：一般情况下，校园的道路是双向通行的，可设校园平面图是一个无向网。顶点和边均含有相关信息。
Campus Guide Consulting (1) basic requirements: (1) to design the floor plan of your school campus, which contain points of interest less than 10. To vertices in the graph various attractions, attractions in store name, code, Profile and other information side of said path storage path length, and other related information. ② any attractions for the visiting guests ask for directions queries, queries between any two of the attractions of a shortest simple path. ③ query information of any attractions for the visiting guests. (2) test data: the reader is specified according to the actual situation. (3) to achieve Note: Under normal circumstances, the campus road is a two-way traffic can be set to the campus plan is to network. Vertices and edges contain relevant information. (2012-06-03, Visual C++, 2KB, 下载4次)
基于最短路径的校园导航问题
任务：设计你的学校的平面图，至少包括10个以上的场所，每两个场所间可以有不同的路，且路长也可能不同，找出从任意场所到达另一场所的最佳路径（最短路径）。
基本功能：
设计校园平面图，在校园景点选10个左右景点。以图中顶点表示校园内各景点，存放景点名称、代号、简介等信息；以边表示路径，存放路径长度等有关信息。
为来访客人提供图中任意景点相关信息的查询。
为来访客人提供任意景点的问路查询，即查询任意两个景点之间的一条最短路径。
实现提示：一般情况下，校园的道路是双向通行的，可设计校园平面图是一个无向网。顶点和边均含有相关信息。
基本要求：
图结构保存在文件中，且允许添加节点与路径。
Based on the campus of the shortest path navigation
Task: to design the floor plan of your school, including at least more than 10 places, each of the two sites can have different road, and the long road may be different, find out the best path to reach the other places (minimum from any place path).
Basic functions:
Design a campus plan to choose 10 attractions on campus attractions. To vertices in the graph campus attractions, attractions in store name, code, Profile and other information to the side of the path to store the path length and other relevant information.
Query information of any attractions for the visiting guests.
Any attractions for the visiting guests ask for directions query, that query the shortest path between any two points of interest.
It realize Tip: Under normal circumstances, the campus road is a two-way traffic can design a campus plan to the network. Vertices and edges contain relevant information.
Basic requirements:
Graph structure is stored in a (2012-04-12, Visual C++, 4KB, 下载13次)
1、 给软件提供一个活跃的校内用户ID（随便到校内网论坛找一个）
2、 软件访问这个ID的校内用户的主页，抓取这个用户的好友。
3、 给这个用户留言。(随机发送一条设置好的留言组中的一条)
4、 访问这个用户的好友，并抓取该好友的好友。给这个用户留言。
5、 重复步骤2到步骤4。（用到了数据结构中的广度优先遍历的算法）
1, to software to provide a vibrant campus user ID (casually to the campus network to find a forum) 2, the software to access the ID of the campus user' s home page, grab the user' s friends. 3, giving the user a message. (Random send a good message in the group to set a) 4, to access this user' s friends, and crawl to the friend' s friend. Give the user a message. 5, repeat steps 2 to step 4. (The data structure used in the breadth-first traversal of the algorithm) (2010-01-17, C/C++, 5517KB, 下载2次)
课程设计：
用无向网表示学校的校园景点平面图，图中顶点表示主要景点，
存放景点的编号、名称、简介等信息，图中的边表示景点间的道路，存放路径长度等信息。要求能够回答有关景点介绍、游览路径等问题。游客通过终端可询问：
（1）从某一景点到另一景点的最短路径。
（2）游客从公园进入，选取一条最佳路线。
（3）使游客可以不重复地浏览各景点，最后回到出口（出口就在入口旁边）。
Curriculum design:
Net free to express School campus sites floor plans, maps vertex express the main attractions,
Store the number of spots, names, information such as brief introduction, the figure that the edge of the road between attractions, such as information storage path length. Requirements be able to answer questions about attractions, the tour path and so on. Tourists through the terminal can be asked:
(1) from one spot to another spot the shortest path.
(2) visitors from the park to enter, select the best route.
(3) so that visitors can browse the various scenic spots in duplicate, and finally back to exports (exports at the entrance on the side). (2009-03-17, C/C++, 227KB, 下载11次)
(本人今年的数据结构课程设计为方便大家特次上传，本站有一个和我的一样，本人曾下过，里面只是部分说明，没有真正的源代码，不知道管理员看没看，那样的作品也能年上传通过)模拟120急救中心响应每个病人的呼救信号统一调度救护车运行的情况。我们对问题作适当简化，假设：某城市共有M个可能的呼救点(居民小区、工厂、学校、公司、机关、单位等)，分布着N所医院(包含在M个点中)，有K辆救护车分派在各医院待命，出现呼救病人时，由急救中心统一指派救护车接送至最近的医院救治。救护车完成一次接送任务后即消毒，并回原处继续待命。假定呼救者与急救中心、急救中心与救护车之间的通讯畅通无阻，也不考虑道路交通堵塞的影响。可以用M个顶点的无向网来表示该城市的各地点和道路。时间可以分钟为单位，路段长可表示为救护车行驶化费的分钟数。
err (2009-01-02, Visual C++, 27KB, 下载14次)
Description 为了宣传本次“网宿科技杯”厦门大学第五届程序设计竞赛，系里面做了两张精美的海报。经过了精确的计算，为了达到最佳美观效果，每张海报都有自己最佳的粘贴位置。但是现在问题是，如果两张海报都要求贴在最佳位置时，很有可能有部分地方会重叠在一起。现在您来判断一下这两张海报是否重叠。 Input 输入包含两行，每一行有四个整数来描述这个海报的最佳粘贴位置，X,Y,W,H(-10000<= X,Y <= 10000)(0 < W, H <= 10000)，X,Y表示海报左下角的坐标，W,H分别表示宽度和高度。 Output 输出"Yes"表示两张海报互相重叠(表示存在一个面积大于0的公共区域)，否则输出"No"。(不包含引号，注意大小写) Sample Input -10 -10 20 30 0 0 30 20 Sample Output Yes (2008-05-18, Visual C++, 1KB, 下载2次)
http://www.pudn.com/Download/item/id/465507.html
掌握Prim算法的特点，学会用Prim算法构造最小生成树
如果无向连通图是一个网，那么它的所有生成树中必有一棵树的边的权值总和为最小，我们称这棵生成树为最小生成树。在Prim算法中，在图G=(V,E)（V表示顶点，E表示边）中任选一点V0，令集合U={V0}为初态，从V0出发寻找与U中顶点相邻（另一顶点在V中）且代价最小的边的另一顶点V1，并使V1加入U，即U={V0，V1}，同时（V0，V1）边加入集合T中（T的初态为空），这样不断地扩大U,直到U=V，则T中即为最小生成树的边。
Grasp the characteristics of Prim algorithm, learning algorithm constructed by Prim minimum spanning tree if the undirected connected graph is a network, then it must be all the spanning tree in the side of a tree weight value sum for the smallest, spanning tree, we called for minimum spanning tree. In the Prim algorithm, in Figure G = (V, E) (V said vertex, E said that edges) in one point V0, the collection U = (V0) for the initial state, starting from V0 to find the adjacent vertex of U (another vertex in V in) and the cost of the other side of the smallest vertex V1, and V1 to join U, namely U = (V0, V1), at the same time (V0, V1) to include in the collection side of T in (T of the initial state is empty), this continues to expand U, until U = V, then T is the minimum spanning tree in the side. (2008-01-11, Visual C++, 1KB, 下载5次)
(1)输入E条弧<j,k>,建立AOE-网的存储结构 (2)从源点v出发,令ve[0]=0,按拓扑排序求其余各项顶点的最早发生时间ve[i](1<=i<=n-1).如果得到的拓朴有序序列中顶点个数小于网中顶点数n,则说明网中存在环,不能求关键路径,算法终止 否则执行步骤(3)(3)从汇点v出发,令vl[n-1]=ve[n-1],按逆拓朴排序求其余各顶点的最迟发生时间vl[i](n-2>=i>=2). (4)根据各顶点的ve和vl值,求每条弧s的最早发生时间e(s)和最迟开始时间l(s).若某条弧满足条件e(s)=l(s),则为关键活动.
(1) E importation of Arc lt; J, kgt; Establish AOE- network storage structure (2) v starting point source, ve [0] = 0, by topological sorting point for the rest of the earliest timing ve [i] (1LT ; = ilt; = n-1). if the Topography vertex orderly sequence number is less than net n vertices, a statement that net presence in Central, not for Critical Path, algorithm implementation steps to terminate or (3) (3) from the Department of v starting point, Vl [n-1] = ve [n-1], by the inverse order for the remaining topology of the latest occurrence of peak time Vl [i] (n-inversion; = IGT; = 2). (4) According to the apex of ve and Vl value for each s arc of the earliest timing e (s) and the latest starting time of l (s). If any meet the conditions of the arc e (s) = l (s), was critical activities. (2005-04-14, C/C++, 2KB, 下载62次)