如何使用 BFS 找到最近的节点？

Question

设G(V,E)为无向未加权图，r为V的子集。现在节点 root 被添加到 G 并且边被添加到 root 和 r 的所有节点之间。现在，对于 V-r 的每个节点，我想使用 BFS 找到最近的 r 节点。请帮忙。我试过下面的代码。

import networkx as nx
import matplotlib.pyplot as plt

def bfs(g, node):
    dist = 0
    visited = [node]
    queue = [(node, dist)]
    tr = {}
    while queue:
        s, dist = queue.pop(0)
        tr[s] = []
        for nbr in list(g.adj[s]):
            if nbr not in visited:
                visited.append(nbr)
                tr[s].append(nbr, dist+1)
                queue.append((nbr, dist+1))
    return tr

G=nx.erdos_renyi_graph(50,0.1)

r=[5,8,36,43,21]
G.add_node('root')
for i in r:
    G.add_edge(i,'root')

t = bfs(G, 'root')
print(t)

Answer 1

我没有理解这个问题中root节点的用途。我认为，解决此问题的最简单方法是，从所有 V-r 节点调用 bfs。每次，当您能够到达属于 r 的节点时，您就 return 它。因为它是属于r的第一个可达节点。这是此过程的示例：

import networkx as nx
import matplotlib.pyplot as plt

def bfs(g, node, r):
    dist = 0
    visited = [node]
    queue = [(node, dist)]
    #tr = {}
    while queue:
        s, dist = queue.pop(0)
        #tr[s] = []
        for nbr in list(g.adj[s]):
            if nbr not in visited:
                if nbr in r:
                    return (nbr, dist+1)
                
                visited.append(nbr)
                #tr[s].append((nbr, dist+1))
                queue.append((nbr, dist+1))
    #return tr
    return (NaN, NaN)

G=nx.erdos_renyi_graph(50,0.1)

r=[5,8,36,43,21]
G.add_node('root')
for i in r:
    G.add_edge(i,'root')

for n in list(G.nodes):
    if n not in r:
        t, d = bfs(G, n, r)
        print("Node {}'s nearest node in r: {} with distance: {}".format(n, t, d))

由于erdos_renyi是一个随机图，所以在不同的运行中可以给出不同的结果。这是一个示例输出：

Node 0's nearest node in r: 5 with distance: 2
Node 1's nearest node in r: 8 with distance: 1
Node 2's nearest node in r: 43 with distance: 2
Node 3's nearest node in r: 36 with distance: 2
Node 4's nearest node in r: 36 with distance: 2
Node 6's nearest node in r: 5 with distance: 2
Node 7's nearest node in r: 36 with distance: 2
Node 9's nearest node in r: 36 with distance: 1
Node 10's nearest node in r: 36 with distance: 2
Node 11's nearest node in r: 8 with distance: 2
Node 12's nearest node in r: 8 with distance: 3
Node 13's nearest node in r: 8 with distance: 2
Node 14's nearest node in r: 8 with distance: 2
Node 15's nearest node in r: 36 with distance: 3
Node 16's nearest node in r: 36 with distance: 3
Node 17's nearest node in r: 8 with distance: 1
Node 18's nearest node in r: 8 with distance: 1
Node 19's nearest node in r: 8 with distance: 1
Node 20's nearest node in r: 21 with distance: 1
Node 22's nearest node in r: 36 with distance: 2
Node 23's nearest node in r: 21 with distance: 2
Node 24's nearest node in r: 21 with distance: 1
Node 25's nearest node in r: 5 with distance: 1
Node 26's nearest node in r: 5 with distance: 1
Node 27's nearest node in r: 36 with distance: 3
Node 28's nearest node in r: 36 with distance: 2
Node 29's nearest node in r: 5 with distance: 1
Node 30's nearest node in r: 21 with distance: 1
Node 31's nearest node in r: 43 with distance: 1
Node 32's nearest node in r: 36 with distance: 3
Node 33's nearest node in r: 5 with distance: 2
Node 34's nearest node in r: 8 with distance: 2
Node 35's nearest node in r: 36 with distance: 2
Node 37's nearest node in r: 36 with distance: 3
Node 38's nearest node in r: 43 with distance: 1
Node 39's nearest node in r: 8 with distance: 2
Node 40's nearest node in r: 43 with distance: 1
Node 41's nearest node in r: 43 with distance: 2
Node 42's nearest node in r: 8 with distance: 2
Node 44's nearest node in r: 43 with distance: 2
Node 45's nearest node in r: 43 with distance: 2
Node 46's nearest node in r: 5 with distance: 2
Node 47's nearest node in r: 8 with distance: 1
Node 48's nearest node in r: 36 with distance: 1
Node 49's nearest node in r: 5 with distance: 2
Node root's nearest node in r: 5 with distance: 1

您甚至可以进一步优化此解决方案。首先，为 V-r 个节点创建一个列表，它将存储从 r 个节点到达该节点的最短距离。用一些大的（即无限的）值初始化这个列表。现在，您可以从所有 r 节点调用 bfs 并在可能的情况下更新距离列表，而不是为每个 V-r 节点调用 bfs。通过这个过程，如果 len(r) << len(V-r)，你将减少对 bfs 的调用。我希望这能解决你的问题。