拓扑图_路径查询

graph = {'A': ['B', 'C','D'],
         'B': [ 'E'],
         'C': ['D','F'],
         'D': ['B','E','G'],
         'E': [],
         'F': ['D','G'],
         'G': ['E']}

import networkx as nx
import numpy as np
import matplotlib.pyplot as plt

G = nx.DiGraph()
G.add_edges_from(
    [('A', 'B'), ('A', 'C'),('A','D'),('B','E'),('C','D'),('C','F'),('D','B'),('D','E'),('D','G'),('F','D'),('F','G'),('G','E')])

nx.draw(G,with_labels=True)
plt.show()

# 找到一条从start到end的路径
def findPath(graph,start,end,path=[]):   
    path = path + [start]
    if start == end:
        return path 
    for node in graph[start]:
        if node not in path:
            newpath = findPath(graph,node,end,path)
            if newpath:
                return newpath
    return None
 
# 找到所有从start到end的路径
def findAllPath(graph,start,end,path=[]):
    path = path +[start]
    if start == end:
        return [path]
 
    paths = [] #存储所有路径    
    for node in graph[start]:
        if node not in path:
            newpaths = findAllPath(graph,node,end,path) 
            for newpath in newpaths:
                paths.append(newpath)
    return paths
 
# 查找最短路径
def findShortestPath(graph,start,end,path=[]):
    path = path +[start]
    if start == end:
        return path
    
    shortestPath = []
    for node in graph[start]:
        if node not in path:
            newpath = findShortestPath(graph,node,end,path)
            if newpath:
                if not shortestPath or len(newpath)<len(shortestPath):
                    shortestPath = newpath
    return shortestPath
 

onepath = findPath(graph,'A','G')
print('一条路径:',onepath)
 
allpath = findAllPath(graph,'A','G')
print('\n所有路径：',allpath)
 
shortpath = findShortestPath(graph,'A','G')
print('\n最短路径：',shortpath)

一条路径: ['A', 'C', 'D', 'G']

所有路径： [['A', 'C', 'D', 'G'], ['A', 'C', 'F', 'D', 'G'], ['A', 'C', 'F', 'G'], ['A', 'D', 'G']]

最短路径： ['A', 'D', 'G']

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Mikowoo007 发布了214 篇原创文章 · 获赞 21 · 访问量 4万+ 私信 关注

标签：node,newpath,end,拓扑图,graph,路径,查询,start,path
来源： https://blog.csdn.net/Mikowoo007/article/details/104464729