问题描述
令 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 并且边被添加在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。我已经尝试了以下代码。
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)
解决方法
我不明白此问题中 root 节点的用途。我认为，解决这个问题的最简单方法是，从所有 bfs 节点调用 V-r。每次，当您能够到达属于 r 的节点时，您将其返回。因为它是属于 r 的第一个可达节点。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))
由于 erdos_renyi 是一个随机图，所以它在不同的运行中会给出不同的结果。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 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。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 bfs 节点调用 bfs 并尽可能更新距离列表，而不是为每个 V-r 节点调用 r。通过此过程，如果 bfs，您将减少对 len(r) << len(V-r) 的调用。我希望这能解决您的问题。