Python中边列表的生成树列表

Question

I am trying to figure out how to print a Spanning Tree list from a given list of edges. For example, if I read in:

0 1

2 1

0 2

1 3

I want to print out a Spanning Tree list of:

[[1], [0,2,3], [1], [1]]

I know how to create an adjacency list using the code:

n = int(input("Enter number of vertices: "))
adjList = [[] for i in range(n)]
with open("graph.txt") as edges:
    for line in edges:
        line = line.replace("\n", "").split(" ")
        adjList[int(line[0])].append(int(line[1]))
        adjList[int(line[1])].append(int(line[0]))
    print(l)

But creating a Spanning Tree is a different story. Given that the Spanning Tree would be unweighted, I am not sure if I need to use some version of Prim's Algorithm here?

Any help is appreciated!

Answer 1

This implementation is based on the networkx package (documentation for it is here ). Note that there's a few different spanning trees that I think you can get for that set of connections. This code will represent the spanning tree by its collection of edges, but I'll see if I can modify it to represent the tree the way you prefer.

import networkx as nx

def spanning_tree_from_edges(edges):
    graph = nx.Graph()

    for n1, n2 in edges:
        graph.add_edge(n1, n2)

    spanning_tree = nx.minimum_spanning_tree(graph)
    return spanning_tree

if __name__ == '__main__':
    edges = [(0, 1), (2, 1), (0, 2), (1, 3)]
    tree = spanning_tree_from_edges(edges)
    print(sorted(tree.edges()))

Output

[(0, 1), (0, 2), (1, 3)]

---

This alternative seems to represent the tree as the collection of node connections (ie in the format you want). Note that this collection is different because it's a different spanning tree to what you get:

if __name__ == '__main__':
    edges = [(0, 1), (2, 1), (0, 2), (1, 3)]
    tree = spanning_tree_from_edges(edges)
    print([list(nx.all_neighbors(tree, n)) for n in tree.nodes()])

Output

[[1, 2], [0, 3], [0], [1]]

Starting Graph

Generated Spanning Tree