Consider a problem to find the shortest path from a given starting node s to any other nodes in the (undirected) graph. The length of a path in the graph is the number of edges included. Complete the function shortest_path(adj_list, s), which takes the adjacency list of the graph and a starting node as input, and returns a list containing the length of the shortest path from the starting node to all other nodes. Note: If there is no path from the starting node to a node, set the length of shortest path as float('inf'). You can use any Python built-in package to implement the stack/queue/priority queue if needed. Hint: Consider the three graph traversal strategies (DFS, BFS and PFS) discussed in the class. Which one is applicable in this problem? For example: Test Result adj_list = [[], [2, 3], [1, 4], [1], [2]] d = shortest_path(adj_list, 0) print(d) d = shortest_path(adj_list, 2) print(d) [0, inf, inf, inf, inf] [inf, 1, 0, 2, 1]

时间: 2024-01-07 09:03:10 浏览: 44
To find the shortest path from a given starting node s to any other nodes in the (undirected) graph, we can use Breadth-First Search (BFS) algorithm. The basic idea of BFS is to explore all the vertices at distance 1 from the current vertex before moving on to vertices at distance 2. Here is the implementation of the shortest_path function using BFS algorithm: ``` from collections import deque def shortest_path(adj_list, s): n = len(adj_list) visited = [False] * n distance = [float('inf')] * n distance[s] = 0 queue = deque([s]) while queue: u = queue.popleft() visited[u] = True for v in adj_list[u]: if not visited[v]: visited[v] = True distance[v] = distance[u] + 1 queue.append(v) for i in range(n): if not visited[i]: distance[i] = float('inf') return distance ``` In the above code, we first initialize two lists: visited and distance. The visited list is used to keep track of the visited nodes and the distance list is used to store the shortest distance from the starting node to all other nodes. We initialize all the distances as infinity except the distance of the starting node which is set to 0. We then use a deque (double-ended queue) to implement the BFS algorithm. We start by adding the starting node to the queue. Then, while the queue is not empty, we remove a vertex u from the front of the queue and mark it as visited. We then iterate over all the neighbors v of u and if v is not visited, we mark it as visited, update its distance from the starting node and add it to the end of the queue. Finally, we check if there are any nodes that were not visited during the BFS traversal and set their distance as infinity. We then return the distance list. Let's use the above code to solve the given example: ``` adj_list = [[], [2, 3], [1, 4], [1], [2]] d = shortest_path(adj_list, 0) print(d) # Output: [0, inf, inf, inf, inf] d = shortest_path(adj_list, 2) print(d) # Output: [inf, 1, 0, 2, 1] ``` In the first test case, the starting node is 0 and there are no edges connected to it. Hence, the distance to all other nodes is infinity. In the second test case, the starting node is 2 and the shortest path to node 2 is 0 (itself). The shortest path to node 1 is 1 (through node 2), the shortest path to node 3 is 2 (through nodes 2 and 1), and the shortest path to node 4 is 1 (through node 2).

相关推荐

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L). To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river. Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N). FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks. Input Line 1: Three space-separated integers: L, N, and M Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position. Output Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks Sample Inputcopy Outputcopy 25 5 2 2 14 11 21 17 4 Hint Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).

最新推荐

recommend-type

Trajectory modification considering dynamic constraints of autonomous robots.pdf

The classic "elastic band" deforms a path generated by a global planner with respect to the shortest path length while avoiding contact with obstacles. It does not take any dynamic constraints of the ...
recommend-type

一种基于A* 算法的动态多路径规划算法

结合一种动态行程时间表对传统A*算法进行调整,可以有效利用路网实时交通数据规避拥堵路线,从而实现动态路径规划。另外,实际应用中,单一的优化路径往往不能满足需求,对此提出重复路径惩罚因子的概念,构造出了一...
recommend-type

智能制造的数字化工厂规划qytp.pptx

智能制造的数字化工厂规划qytp.pptx
recommend-type

zigbee-cluster-library-specification

最新的zigbee-cluster-library-specification说明文档。
recommend-type

管理建模和仿真的文件

管理Boualem Benatallah引用此版本:布阿利姆·贝纳塔拉。管理建模和仿真。约瑟夫-傅立叶大学-格勒诺布尔第一大学,1996年。法语。NNT:电话:00345357HAL ID:电话:00345357https://theses.hal.science/tel-003453572008年12月9日提交HAL是一个多学科的开放存取档案馆,用于存放和传播科学研究论文,无论它们是否被公开。论文可以来自法国或国外的教学和研究机构,也可以来自公共或私人研究中心。L’archive ouverte pluridisciplinaire
recommend-type

实现实时数据湖架构:Kafka与Hive集成

![实现实时数据湖架构:Kafka与Hive集成](https://img-blog.csdnimg.cn/img_convert/10eb2e6972b3b6086286fc64c0b3ee41.jpeg) # 1. 实时数据湖架构概述** 实时数据湖是一种现代数据管理架构,它允许企业以低延迟的方式收集、存储和处理大量数据。与传统数据仓库不同,实时数据湖不依赖于预先定义的模式,而是采用灵活的架构,可以处理各种数据类型和格式。这种架构为企业提供了以下优势: - **实时洞察:**实时数据湖允许企业访问最新的数据,从而做出更明智的决策。 - **数据民主化:**实时数据湖使各种利益相关者都可
recommend-type

解释minorization-maximization (MM) algorithm,并给出matlab代码编写的例子

Minorization-maximization (MM) algorithm是一种常用的优化算法,用于求解非凸问题或含有约束的优化问题。该算法的基本思想是通过构造一个凸下界函数来逼近原问题,然后通过求解凸下界函数的最优解来逼近原问题的最优解。具体步骤如下: 1. 初始化参数 $\theta_0$,设 $k=0$; 2. 构造一个凸下界函数 $Q(\theta|\theta_k)$,使其满足 $Q(\theta_k|\theta_k)=f(\theta_k)$; 3. 求解 $Q(\theta|\theta_k)$ 的最优值 $\theta_{k+1}=\arg\min_\theta Q(
recommend-type

JSBSim Reference Manual

JSBSim参考手册,其中包含JSBSim简介,JSBSim配置文件xml的编写语法,编程手册以及一些应用实例等。其中有部分内容还没有写完,估计有生之年很难看到完整版了,但是内容还是很有参考价值的。
recommend-type

"互动学习:行动中的多样性与论文攻读经历"

多样性她- 事实上SCI NCES你的时间表ECOLEDO C Tora SC和NCESPOUR l’Ingén学习互动,互动学习以行动为中心的强化学习学会互动,互动学习,以行动为中心的强化学习计算机科学博士论文于2021年9月28日在Villeneuve d'Asq公开支持马修·瑟林评审团主席法布里斯·勒菲弗尔阿维尼翁大学教授论文指导奥利维尔·皮耶昆谷歌研究教授:智囊团论文联合主任菲利普·普雷教授,大学。里尔/CRISTAL/因里亚报告员奥利维耶·西格德索邦大学报告员卢多维奇·德诺耶教授,Facebook /索邦大学审查员越南圣迈IMT Atlantic高级讲师邀请弗洛里安·斯特鲁布博士,Deepmind对于那些及时看到自己错误的人...3谢谢你首先,我要感谢我的两位博士生导师Olivier和Philippe。奥利维尔,"站在巨人的肩膀上"这句话对你来说完全有意义了。从科学上讲,你知道在这篇论文的(许多)错误中,你是我可以依
recommend-type

实现实时监控告警系统:Kafka与Grafana整合

![实现实时监控告警系统:Kafka与Grafana整合](https://imgconvert.csdnimg.cn/aHR0cHM6Ly9tbWJpei5xcGljLmNuL21tYml6X2pwZy9BVldpY3ladXVDbEZpY1pLWmw2bUVaWXFUcEdLT1VDdkxRSmQxZXB5R1lxaWNlUjA2c0hFek5Qc3FyRktudFF1VDMxQVl3QTRXV2lhSWFRMEFRc0I1cW1ZOGcvNjQw?x-oss-process=image/format,png) # 1.1 Kafka集群架构 Kafka集群由多个称为代理的服务器组成,这