如何使用BFS查找最近的节点？

Question

设G(V,E)为无向无权图r为V的子集。现在，节点root被添加到G，并且在root和r的所有节点之间添加边。现在，对于V-r的每个节点，我想使用BFS查找最近的r节点。请帮忙。我尝试了以下代码

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

我不理解root节点在这个问题中的作用。我认为，解决这个问题最简单的方法是，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，都会返回它。因为它是属于r的第一个可到达节点。以下是此过程的示例：

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))

因为erdos_renyi是一个随机图，所以它可以在不同的运行中给出不同的结果。以下是一个示例输出：

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节点创建一个列表，该列表将存储从r节点到达该节点的最短距离。使用一些较大的（即无限）值初始化此列表。现在，不必为每个V-r节点调用bfs，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。通过这个过程，如果len(r) << len(V-r)，您对bfs的调用将更少。我希望这能解决你的问题