如何在广度优先树中存储节点到根的距离？

问题描述

令 G(V,E) 为无向图。我想从 G 存储与根具有相应距离的节点创建一个广度优先树。请帮忙。

import networkx as nx

g=nx.erdos_renyi_graph(12,.1)
visited = [] 
queue = []     
tree={}
def bfs(visited,g,node):
  visited.append(node)
  queue.append(node)

  while queue:
    s = queue.pop(0)
    tree[s]=[]
    for neighbour in list(g.adj[s]):
      if neighbour not in visited:
        visited.append(neighbour)
        tree[s].append(neighbour)
        queue.append(neighbour)

bfs(visited,1)
print(tree)

解决方法

您可以将每个项目保留为 (node,distance) 并附加 (neighbour,distance+1)

import matplotlib.pyplot as plt
import networkx as nx

# --- functions ---

def bfs(g,node):
    distance = 0
    visited = [node]
    queue = [(node,distance)]

    tree = {}

    while queue:
        s,distance = queue.pop(0)
        tree[s] = []
        for neighbour in list(g.adj[s]):
            if neighbour not in visited:
                visited.append(neighbour)
                tree[s].append((neighbour,distance+1))
                queue.append((neighbour,distance+1))

    return tree
  
# --- main ---

#G = nx.erdos_renyi_graph(12,.1)
G = nx.Graph()
G.add_edges_from(([1,2],[2,3],4],[3,5],[4,5]))

tree = bfs(G,1)
print(tree)

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

结果（手动重整）：

{
  1: [(2,1)],2: [(3,2),(4,2)],3: [(5,3)],4: [],5: []
}

对于图表

---

最终你甚至可以使用 tree[(s,distance)]

def bfs(g,distance = queue.pop(0)
        tree[(s,distance)] = []
        for neighbour in list(g.adj[s]):
            if neighbour not in visited:
                visited.append(neighbour)
                tree[(s,distance)].append((neighbour,distance+1))

    return tree

结果（手动重整）：

{
  (1,0): [(2,(2,1): [(3,(3,2): [(5,2): [],(5,3): []
}

---

顺便说一句：

我想使用 json.dumps(tree,indent=2) 自动重新格式化结果，但它无法将 node 转换为 json。

但是 pretty printer 可以用类似的方式格式化

import pprint
pprint.pprint(tree,indent=2)

结果：

{ (1,3): []}