你可以使用
defaultdict
从边/路径列表创建你的“图形”:
edges = [['1', '2'], ['2', '4'], ['1', '11'], ['4', '11']]
G = defaultdict(list)
for (s,t) in edges:
G[s].append(t)
G[t].append(s)
print G.items()
输出:
[
('1',['2','11']),
('11',['1','4']),
('2',['1','4']),
('4',['2','11'])
]
请注意,我添加了双向边,因为您正在使用无向图。因此,对于边(a,b),G [a]
将包括b
,而G [b]
将包括a
。
从这里,您可以使用像深度优先搜索或广度优先搜索这样的算法来发现图中的所有路径。
在下面的代码中,我使用了DFS:
def DFS(G,v,seen=None,path=None):
if seen is None: seen = []
if path is None: path = [v]
seen.append(v)
paths = []
for t in G[v]:
if t not in seen:
t_path = path + [t]
paths.append(tuple(t_path))
paths.extend(DFS(G, t, seen[:], t_path))
return paths
以下是您可以使用的内容:
G = defaultdict(list)
for (s,t) in edges:
G[s].append(t)
G[t].append(s)
print DFS(G, '1')
输出:
[('1', '2'), ('1', '2', '4'), ('1', '2', '4', '11'), ('1', '11'), ('1', '11', '4'), ('1', '11', '4', '2')]
因此,完整的代码,包括显示最长路径的最后一部分:
from collections import defaultdict
def DFS(G,v,seen=None,path=None):
if seen is None: seen = []
if path is None: path = [v]
seen.append(v)
paths = []
for t in G[v]:
if t not in seen:
t_path = path + [t]
paths.append(tuple(t_path))
paths.extend(DFS(G, t, seen[:], t_path))
return paths
edges = [['1', '2'], ['2', '4'], ['1', '11'], ['4', '11']]
G = defaultdict(list)
for (s,t) in edges:
G[s].append(t)
G[t].append(s)
all_paths = DFS(G, '1')
max_len = max(len(p) for p in all_paths)
max_paths = [p for p in all_paths if len(p) == max_len]
print("All Paths:")
print(all_paths)
print("Longest Paths:")
for p in max_paths: print(" ", p)
print("Longest Path Length:")
print(max_len)
输出:
所有路径:
[('1', '2'), ('1', '2', '4'), ('1', '2', '4', '11'), ('1', '11'), ('1', '11', '4'), ('1', '11', '4', '2')]
最长路径:
('1', '2', '4', '11')
('1', '11', '4', '2')
最长路径长度:
4
请注意,您搜索的“起点”由DFS
函数的第二个参数指定,在本例中为'1'
。
更新:如评论中所讨论的,上述代码假设您有一个起点(具体来说,代码使用标记为'1'
的节点)。
更一般的方法是,在没有这样的起点的情况下,从每个节点开始执行搜索,并取最长路径。(注意:实际上,您可以比这更聪明)
更改行:
all_paths = DFS(G, '1')
为了
all_paths = [p for ps in [DFS(G, n) for n in set(G)] for p in ps]
会给出任意两点之间最长的路径。
(这是一个愚蠢的列表推导式,但它允许我只更新一行。更清楚地说,它等价于以下内容:
all_paths = []
for node in set(G.keys()):
for path in DFS(G, node):
all_paths.append(path)
或者
from itertools import chain
all_paths = list(chain.from_iterable(DFS(G, n) for n in set(G)))
).