我有一个用networkx开发的python G有向图。该图具有称为“权重”的权重。
我知道一个显式的起始节点 A 和一个结束节点 F。在图之间可以访问节点 B、C、D、E。
我怎么能明确地说他必须通过找到最短路径来访问B和D，并且可以另外添加C和E，如果这有助于最短路径？
到目前为止我知道这个功能:
nx.single_source_dijkstra(G, 'A', target='F', cutoff=None, weight='weight')
给出输出:
(10.01,
['A',
'B',
'C',
'F',])
如何确保它包含 E？
最佳答案
Networkx 没有内置函数或参数来解决您的问题。您应该手动执行此操作:
import networkx as nx
# Create a random DAG
G = nx.gnp_random_graph(50,0.3,directed=True)
DAG = nx.DiGraph([(u,v) for (u,v) in G.edges() if u<v])
nx.is_directed_acyclic_graph(DAG)
for edge in G.edges:
G.edges[edge]['weight'] = 1
# Get the longest path (without weights) from node 1 to node 40
# with nodes 5, 10, 20, 30 inside
max([
(path, len(path))
for path in nx.all_simple_paths(DAG, 1, 40)
if all(n in path for n in (5, 10, 20, 30))
], key=lambda x: x[1])
# Get the longest path (with weights)
max([
path
for path in nx.all_simple_paths(DAG, 1, 40)
if all(n in path for n in (5, 10, 20, 30))
], key=lambda x: sum(G.edges[edge]['weight'] for edge in nx.utils.pairwise(x)))
关于python - networkx 中具有 'key' 个节点可访问的最短路径，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/56460584/