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[Data structs] 数据结构车务管理 DL : 0: 用数据结构实现基本算法用C++实现车务管理，需用到栈基本数据结构-data structure used to achieve basic algorithm to achieve C bus services management, needed to stack the basic data structure
Date : 2025-12-29 Size : 130kb User : 孙锋志
[Data structs] 复件轮渡问题 DL : 0: 1．汽车轮渡口，过江渡船每次能载10辆车过江，过江车辆分为客车类和货车类，上渡船有如下规定：同类车先到先上船，客车先于货车上渡船，且每上4辆客车，才允许上一辆货车。若等待客车不足4辆，则从货车代替，若无货车等待允许客车上船。试写一个算法模拟渡口管理。算法设计: 1客车和货车均建立一个链式队列，初始均为空。以后来一辆车不是货车就是客车，因此可以说整个程序的事件驱动event就是这两个,客车表示1，货车表示0. 2轮船还没有到达时客车和货车均按次序排在各自队列中。 3轮船到达时，根据两个队列的情况，分别处理。处理如下： a 客车数不满4辆，则将排在前面的货车上船，但总数不能超过10，若没有货车等待，客车直接上船。 b 客车数满4，但不满8辆，客车先上，排在前面的只有一辆货车可以上船，若没有货车等待则货车不上。 c 客车满8辆但不满10，客车上船，排在前面的货车最多可以上2辆，但总数不能超过10。 d 客车满10，则全上客车,但总数不能超过10。 -1. I car ferry, crossing the river each ferry can carry 10 cars crossing the river, crossing the river into passenger vehicles and goods category, Ferry on the following provisions : first vehicle in a first embarkation, the first passenger ferry in the truck and four on each passenger will be permitted on a lorry. If waiting for the bus less than four, then replace the truck, without waiting for the lorry to allow passenger embarkation. Try to write a simulated crossing management. Algorithm design : a bus and the truck were established a chain cohort, the initial were empty. Later, a car is not a passenger vehicle is, it can be said of the entire process event-driven event is the two, said a passenger, said the lorry 0. Two ships have not yet arrived at the bus and the truck were ranked
Date : 2025-12-29 Size : 2kb User : 西们子
[Data structs] ferry DL : 0: 该算法模拟渡口管理。过江渡船每次能载10辆车过江，过江车辆分为客车类和货车类，上渡船有如下规定：同类车先到先上船，客车先于货车上渡船，且每上4辆客车，才允许上一辆货车。若等待客车不足4辆，则从货车代替，若无货车等待允许客车上船。-the algorithm simulation ferry management. Jiang Ferry each can carry 10 cars crossing the river, crossing the river into passenger vehicles and goods category, Ferry on the following provisions : Similar cars first come-first embarkation, the first passenger ferry in the truck and four on each passenger will be permitted on a lorry. If waiting for the bus less than four, then replace the truck, without waiting for the lorry to allow passenger embarkation.
Date : 2025-12-29 Size : 1kb User : Ash
[Data structs] Neobus0.1-2 DL : 0: 北京公交车查询系统北京公交车查询系统-Beijing Bus Inquiry System Beijing Bus Inquiry System
Date : 2025-12-29 Size : 616kb User : 周华
[Data structs] lujing DL : 0: 公车最短路径算法具体的算法：实际上乘客不一定要找最短，里面还涉及到换乘次数最少，费用最少等问题因此，边向量的权值要分多种情况考虑，所以如果真的要开发实用的系统应该给出多种最佳选择。 -Bus shortest path algorithm specific algorithm: in fact they do not necessarily find the shortest, which also involves the least number of transfer cost and other issues at least, therefore, the right edge vector values have to consider a number of situations, so if we really wish to the development of practical systems should be given a variety of best choice.
Date : 2025-12-29 Size : 4kb User : fung
[Data structs] dukou DL : 0: 有一个渡口，每条渡轮一次能装载10辆汽车过江，过江车辆分为客车和货车两类，上渡轮有如下规定： ⑴同类汽车先到先上船； ⑵客车先于货车上船； ⑶每上4辆客车才允许上一辆货车，但若等待的客车不足4辆则用货车填补，反过来，若没有货车等待则用客车填补； ⑷装满10辆后则自动开船，当等待时间较长时车辆不足10辆也应人为控制发船。 -There is a ferry, and each time the ferry can carry 10 vehicles crossing the river, crossing the river of vehicles is divided into two types of passenger cars and trucks, on the ferry the following provisions: ⑴ similar car first-come, first embarkation ⑵ bus before the goods on board ⑶ every passenger on the four will be allowed to board a goods vehicle, but if waiting for a bus less than four are filled by trucks, in turn, if there is no vehicle waiting for the bus to fill with ⑷ filled automatically after 10 set sail, when the a longer waiting time less than 10 when the vehicle should also be man-made ship-fat control.
Date : 2025-12-29 Size : 260kb User : 张巨松
[Data structs] gjcx DL : 0: 本程序实现的是南京的公交线路查询，可以实现任意两个站点的最优乘车路线显示-This procedure is to realize the bus lines Nanjing query, you can realize any two sites showed that the optimal route by car
Date : 2025-12-29 Size : 102kb User : cjc
[Data structs] df DL : 0: 一．需求分析 1. 本程序的功能接受用户的中文输入，用string 保存，能够按选定的数据结构设计相应的算法。当从乘车站到目的站存在多种乘车路线时，可以确定选取标准。例如，要求换车次数最少、经过站点最少等。 2. 输入形式输入站名或本系统提供的站名所对应的序号 3. 输出的形式输出按用户的选取标准的路线 -1. Demand Analysis 1. This procedure function of the Chinese to accept user input, using string preservation, according to the selected data structure design corresponding algorithms. When traveling from the station to the destination station when there are multiple bus routes, to determine selection criteria. For example, the least number of requests to change, after a minimum of such sites. 2. Enter the station name or the form of input provided by the system corresponding to station No. 3. Output form the output selected by the user
Date : 2025-12-29 Size : 6kb User :
[Data structs] m-in DL : 0: 数据隐私保护K-匿名问题的动态多次发布类似m-invarience算法-Data Privacy Protection K-anonymity problem on many occasions issued a similar dynamic m-invarience algorithm
Date : 2025-12-29 Size : 6kb User : 王冠
[Data structs] b DL : 0: 【问题描述】: 有一个渡口，每条渡轮一次能装载10辆汽车过江，过江车辆分为客车和货车两类，上渡轮有如下规定： ⑴同类汽车先到先上船； ⑵客车先于货车上船； ⑶每上4辆客车才允许上一辆货车，但若等待的客车不足4辆则用货车填补，反过来，若没有货车等待则用客车填补； ⑷装满10辆后则自动开船，当等待时间较长时车辆不足10辆也应人为控制发船。【实现提示】：此题可建立和使用两个链式队列，一个为客车队列，另一个为货车队列，到渡口需过江的汽车分别进入到相应队列中。当渡口有渡轮时先让客车队列中的4部客车出队并开进渡轮，再让货车队列中的4部货车出队并开进渡轮，若某一类车辆队列为空则从另一队列中补充。当渡轮上装满10辆后则自动开船，此时应输出已装每辆车的车号。若装载不足10辆，但两个车辆队列全为空，应继续等待一段时间，若等待时间较长，仍不满载则应人为控制开船。 -Description of the problem: There is a ferry, and each ferry will carry a 10 car crossing the river, the river traffic is divided into two types of passenger cars and trucks, on the ferry the following provisions: ⑴ similar first-come, first car on board ⑵ bus before the goods on board ⑶ 4 per passenger on board a goods vehicle will be allowed, but waiting for the bus less than 4 are filled with goods, in turn, if there is no vehicle waiting for the bus to fill with ⑷ filled automatically after 10 set sail, when the waiting time less than 10 when the vehicle should also be made artificially control the ship. 【Achieve prompt: This title can be established and the use of two chain queue, a queue for the bus, and another queue for the goods, to take the ferry crossing the river into the car, respectively, the corresponding queue. When there are ferry crossing bus queue when the first four of the team bus and into the ferry, the queue of trucks allowed four trucks and
Date : 2025-12-29 Size : 5kb User : pmy
[Data structs] 123456789 DL : 0: 最短路径计算,计算两点之间最短的公交线路-The shortest path calculation between two points the shortest bus route
Date : 2025-12-29 Size : 3kb User : 开机
[Data structs] bus_stop DL : 0: 停车场管理系统,用队列和栈实现车辆管理,记录车辆到达和离开时间-bus stop system,using stack and ...
Date : 2025-12-29 Size : 269kb User : 杨旸
[Data structs] BUS_BIN DL : 0: (estructura de datos) programa en C++ para ejemplificar bus binario
Date : 2025-12-29 Size : 1kb User : jim
[Data structs] Bus-inquires-the-idea DL : 0: 公交换乘设计思路，实现最多两次换乘。公交换乘思路大概是这样，当然还可以使用二叉树-Bus transfer design ideas to achieve a maximum of two transfer. Bus transfer thinking about this, of course, you can also use the binary tree
Date : 2025-12-29 Size : 1kb User : uc
[Data structs] OS_driver-and-ticket DL : 0: 要求完成的主要任务: （包括课程设计工作量及其技术要求，以及说明书撰写等具体要求） 1．模拟公共汽车司机和售票员开关门及行车操作的同步模型。 2．设计报告内容应说明： ⑴ 需求分析； ⑵ 功能设计（数据结构及模块说明）； ⑶ 开发平台及源程序的主要部分； ⑷ 测试用例，运行结果与运行情况分析； ⑸ 自我评价与总结 -Required to complete major tasks: (including course design work and technical requirements, as well as written instructions and other specific requirements) 1. Analog bus driver and conductor switch doors and traffic operations synchronization model. 2. Design report shall state: ⑴ needs analysis ⑵ functional design (data structure and the module description) ⑶ source development platform and the main part ⑷ test, analysis of operating results and operating conditions ⑸ self-evaluation and summary
Date : 2025-12-29 Size : 899kb User : 张亮
[Data structs] bus DL : 0: 数据结构编程题目，BUS问题，使用C语言编程-data structure programming problem,bus problems,with language
Date : 2025-12-29 Size : 1kb User : huashui
[Data structs] the-Bus-Magement--System DL : 0: 1.查找公交直达路线 2.删除公交路线 3.增加公交路线 4.查询最短公交路线 -1 direct bus route to find 2 Remove the bus route 3 bus routes to increase 4 the shortest transit route query
Date : 2025-12-29 Size : 3kb User : kitty
[Data structs] to-see-the-Olympics-by-bus DL : 0: 数学建模题目-乘公交看奥运代码，实现北京地铁公交换乘次数少，时间少，费用少-Mathematical modeling topics- code by bus to see the Olympics, realization of Beijing subway transit times less, less time and less cost
Date : 2025-12-29 Size : 749kb User : zhoufeifei
[Data structs] see-the-Olympics-by-bus-and-subway DL : 0: 数学建模题目-乘公交和地铁看奥运代码，实现北京地铁公交换乘次数少，时间少，费用少-Mathematical modeling topics- by bus and subway watching Olympic code, to achieve the Beijing subway transit times less, less time and less cost
Date : 2025-12-29 Size : 39kb User : zhoufeifei
[Data structs] Check bus routes DL : 0: 问题描述当一个用户从甲地到乙地时，由于不同需求，就有不同的交通方式及不同的交通路线。有人希望以最快速度到达，有人希望以最短距离到达，有人希望用最少的费用等。交通方式有公交车和地铁。编写一北京公交线路查询系统，通过输入起始站、终点站，为用户提供三种或以上决策的交通咨询。(problem description When a user goes from point a to point b, there are different modes of transportation and different routes due to different demands. Some people want to arrive at the fastest speed, some people want to arrive at the shortest distance, some people want to use the least cost and so on. There are buses and subways. Compile a Beijing bus line inquiry system, provide users with three or more kinds of traffic consultation by entering the starting station and the terminal station.)
Date : 2025-12-29 Size : 421kb User : 想梦

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