minimum-spanning-tree

问题 I am struggling with this. We can get MST using Kruskal's algorithm or Prim's algorithm for the MST. And for "second-best" MST, I can: first get MST using either of the algorithm mentioned above. For each V-1 of the optimal edge from the MST: a. first remove or flag the edge b. continue calculating MST without that edge c. compare and record down that "second-best" MST with previous iteration In the end we have "second-best" MST But this runs in O(VE) where V is num of vertex and E is number

问题 I am trying to implement a randomly generated maze using Prim's algorithm. I want my maze to look like this: however the mazes that I am generating from my program look like this: I'm currently stuck on correctly implementing the steps highlighted in bold: Start with a grid full of walls. Pick a cell, mark it as part of the maze. Add the walls of the cell to the wall list. While there are walls in the list: **1. Pick a random wall from the list. If the cell on the opposite side isn't in the

问题 I've been looking for an implementation (I'm using networkx library.) that will find all the minimum spanning trees (MST) of an undirected weighted graph. I can only find implementations for Kruskal's Algorithm and Prim's Algorithm both of which will only return a single MST. I've seen papers that address this problem (such as Representing all minimum spanning trees with applications to counting and generation) but my head tends to explode someway through trying to think how to translate it

问题 I have written a code that solves MST using Prim method. I read that this kind of implementation(using priority queue) should have O(E + VlogV) = O(VlogV) where E is the number of edges and V number of Edges but when I look at my code it simply doesn't look that way.I would appreciate it if someone could clear this up for me. To me it seems the running time is this: The while loop takes O(E) times(until we go through all the edges) Inside that loop we extract an element from the Q which takes

问题 I have a list as c4_leaves = [56,78,90,112] . I'm trying to create a complete graph using these elements in c4_leaves as nodes. Here's what I've tried; V_ex = c4_leaves G_ex = nx.Graph() G_ex.add_nodes_from(V_ex) G_ex = nx.complete_graph(4) for u,v in G_ex.edges(): G_ex[u][v]['distance'] = distance(points33, u, v) And then the minimum spanning tree of the above graph as : T_ex= nx.minimum_spanning_tree(G_ex, weight='distance') F_ex = list(T_ex.edges()) When I draw G_ex , it gives me the

问题 Given an undirected connected graph with weights. w:E->{1,2,3,4,5,6,7} - meaning there is only 7 weights possible. I need to find a spanning tree using Prim's algorithm in O(n+m) and Kruskal's algorithm in O( m*a(m,n)). I have no idea how to do this and really need some guidance about how the weights can help me in here. 回答1: You can sort edges weights faster. In Kruskal algorithm you don't need O(M lg M) sort, you just can use count sort (or any other O(M) algorithm). So the final complexity

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: All minimum spanning trees implementation How can I find all minimum spanning trees in an undirected graph in an efficient way? 回答1: Apologies for the academic answer... but algorithm S in Knuth's TAOCP, Volume 4, Fascicle 4 is exactly about generating all spanning trees (pp. 26ff). There are a few musings when he talks about generating (spanning) trees, but your best bet in TAOCP. 回答2: you can find one.

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: All minimum spanning trees implementation How can I find all minimum spanning trees in an undirected graph in an efficient way? 回答1: Apologies for the academic answer... but algorithm S in Knuth's TAOCP, Volume 4, Fascicle 4 is exactly about generating all spanning trees (pp. 26ff). There are a few musings when he talks about generating (spanning) trees, but your best bet in TAOCP. 回答2: you can find one.

问题 What is the exact difference between Dijkstra's and Prim's algorithms? I know Prim's will give a MST but the tree generated by Dijkstra will also be a MST. Then what is the exact difference? 回答1: Prim's algorithm constructs a minimum spanning tree for the graph, which is a tree that connects all nodes in the graph and has the least total cost among all trees that connect all the nodes. However, the length of a path between any two nodes in the MST might not be the shortest path between those

问题 I'm trying to implement an algorithm in c which finds MST using DFS. I've already found the DFS algortihm and i understand it pretty well. I also know that i should follow these steps to achieve my purpose : 1 Run DFS till you find an edge going backwards or DFS stopped. If stopped return G. 2 On the circle that is constructed by the backwards going edge find the heaviest edge and remove it from G. 3 Return to 1. Here is the DFS code : #include<stdio.h> void DFS(int); int G[10][10],visited[10