📅  最后修改于: 2023-12-03 15:12:44.242000             🧑  作者: Mango
该问题要求在给定的有向图中找到所谓的门控点,门控点是指当该点被移除后,有向图会被拆分成多个连通组件。
可以采用深度优先搜索(DFS)算法解决该问题。对于给定的有向图,我们需要找到门控点,就需要做到以下几点:
代码如下:
visited = set() # 已经被访问过的节点集合
d = {} # 节点最早发现时间的字典
low = {} # 节点发现的最早祖先节点的字典
parent = {} # 节点的父节点字典
is_gate = {} # 节点是否是门控点的字典
time = 0 # 当前时间
root = None # 根节点
def dfs(u, graph):
global visited, d, low, parent, is_gate, time, root
visited.add(u)
d[u] = time
low[u] = time
time += 1
children_count = 0
for v in graph[u]:
if v not in visited:
parent[v] = u
if u == root:
children_count += 1
dfs(v, graph)
low[u] = min(low[u], low[v])
if low[v] >= d[u] and u != root:
is_gate[u] = True
elif u == root and children_count > 1:
is_gate[u] = True
elif v != parent[u]:
low[u] = min(low[u], d[v])
def find_gates(graph):
global visited, d, low, parent, is_gate, time, root
visited = set()
d = {}
low = {}
parent = {}
is_gate = {}
time = 0
root = next(iter(graph.keys()))
dfs(root, graph)
is_gate[root] = (len(graph[root]) > 1)
return [u for u in is_gate if is_gate[u]]
这篇文章介绍了解决门控点问题的常见方法——深度优先搜索(DFS)。需要注意的是,虽然这种方法是在有向图上实现的,但在无向图上同样适用。