depth-first-search

问题 I've seen other questions about depth first search but my problem is slightly different and I really don't get it. In prolog, I represent my undirected graph like this: [0-[1,5], 1-[0,2], 2-[1,3], 3-[2,0], 5-[0]] Which is a set of key-value, where the key represents the node and the list: -[] represents its neighbors. I don't know how to do a depth first search with this model. I've tried many solution. I want a really basic recursive algorithm like this one: dfs(v, visited): if visited[v]

问题 This question already has answers here : Why is time complexity for BFS/DFS not simply O(E) instead of O(E+V)? (2 answers) Breadth First Search time complexity analysis (6 answers) Closed 9 days ago . This is a generic BFS implementation: For a connected graph with V nodes and E total number of edges, we know that every edge will be considered twice in the inner loop. So if the total number of iterations in the inner loop of BFS is going to be 2 * number of edges E , isn't the runtime going

问题 This question already has answers here : Why is time complexity for BFS/DFS not simply O(E) instead of O(E+V)? (2 answers) Breadth First Search time complexity analysis (6 answers) Closed 9 days ago . This is a generic BFS implementation: For a connected graph with V nodes and E total number of edges, we know that every edge will be considered twice in the inner loop. So if the total number of iterations in the inner loop of BFS is going to be 2 * number of edges E , isn't the runtime going

问题 I know there's a similar question in stack overflow, where one person has asked, why time complexity of BFS/DFS is not simply O(V). The appropriate answer given was that E can be as large as V^2 in case of complete graph, and hence it is valid to include E in time complexity. But, if V cannot be greater than E+1. So, in that case not having V in the time complexity, should work? 回答1: If it is given that E = kV + c , for some real constants k and c then, O(E + V) = O(kV + c + V) = O(V) = O(E)

问题 I know there's a similar question in stack overflow, where one person has asked, why time complexity of BFS/DFS is not simply O(V). The appropriate answer given was that E can be as large as V^2 in case of complete graph, and hence it is valid to include E in time complexity. But, if V cannot be greater than E+1. So, in that case not having V in the time complexity, should work? 回答1: If it is given that E = kV + c , for some real constants k and c then, O(E + V) = O(kV + c + V) = O(V) = O(E)

问题 I am currently using networkx library for Python with BFS and DFS. I need to get a tree and then explore it to get a path from a start node to an end node. For the BFS part I am using bfs_successors and it returns an iterator of successors in breadth-first-search from source. For the DFS part I am using: dfs_successors and it returns a dictionary of successors in depth-first-search from source. I need to get a list of nodes from source to end from both the algorithms. Each node is (x, y) and

问题 I'm having trouble computing the path from the root to a specified node in a binary tree (this is specifically about a Python solution to this problem). Here's an example. Given the binary tree below, if I specify the node whose value is 4, I want to return [1, 2, 4]. If I specify the node whose value is 5, I want to return [1, 2, 5]. 1 / \ 2 3 / \ 4 5 Here's my attemped solution. class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None def path(root, k, l=[]):

问题 I'm having trouble computing the path from the root to a specified node in a binary tree (this is specifically about a Python solution to this problem). Here's an example. Given the binary tree below, if I specify the node whose value is 4, I want to return [1, 2, 4]. If I specify the node whose value is 5, I want to return [1, 2, 5]. 1 / \ 2 3 / \ 4 5 Here's my attemped solution. class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None def path(root, k, l=[]):

问题 Can you please let me know what is incorrect in below DFS code. It's giving correct result AFAIK, but I don't know when it will fail. graph1 = { 'A' : ['B','S'], 'B' : ['A'], 'C' : ['D','E','F','S'], 'D' : ['C'], 'E' : ['C','H'], 'F' : ['C','G'], 'G' : ['F','S'], 'H' : ['E','G'], 'S' : ['A','C','G'] } visited = [] def dfs(graph,node): global visited if node not in visited: visited.append(node) for n in graph[node]: dfs(graph,n) dfs(graph1,'A') print(visited) Output: ['A', 'B', 'S', 'C', 'D',

问题 DFS with coloring would take O(V+E) vs union find would take O(ElogV) reference: http://www.geeksforgeeks.org/detect-cycle-undirected-graph/ So union find approach is slower. If V = 100, E = 100, DFS = 200, Union find is 1,000. Is there a reason to use Union find? I personally like it because it produces a clean code. Or anything I missed that union find is better in real practice? 回答1: I suspect that you may be misinterpreting how big-O notation works. The notation O(V + E) doesn't mean "the