📅  最后修改于: 2023-12-03 15:10:25.470000             🧑  作者: Mango
有向欧拉电路指的是经过图中每条有向边恰好一次的路径。而当我们面对一个无向图,需要将其转化为有向图以满足有向欧拉电路的条件时,我们可以使用以下方法。
首先,在无向图中,我们需要找到所有奇度数节点,将这些节点的度数减一,使得它们的度数变为偶数。因为在无向图中,每个节点的度数为偶数时,我们可以将它们分成两个集合,并在这两个集合之间来回走动,经过每条边恰好一次。
接着,我们可以将每个奇度数节点拆成两个节点,并在这两个节点间连一条有向边,使得图中的每个节点度数均为偶数。这样就将无向图转化为了有向图。
最后,我们可以使用Hierholzer's算法来找到有向欧拉回路。该算法的基本思路是贪心地遍历图中的每条边,如果当前节点的所有出边都已经被访问过了,就将该节点加入到遍历的序列中,然后倒序遍历序列,以此找到有向欧拉回路。
以下为Python的示例代码:
def undirected_to_directed_eulerian_circuit(graph):
# Find all nodes with odd degrees and subtract 1 from their degrees.
for node in graph:
if len(graph[node]) % 2 == 1:
graph[node].append("_dummy")
graph["_dummy"] = [node]
# Create a directed graph by replacing each node with two nodes
# and adding a directed edge between them.
directed_graph = {}
for node in graph:
if len(graph[node]) > 0:
directed_graph[(node, 0)] = [(neighbor, 1) for neighbor in graph[node]]
directed_graph[(node, 1)] = []
else:
directed_graph[(node, 0)] = []
# Find an Eulerian circuit in the directed graph.
circuit = []
stack = [(next(iter(directed_graph)), None)]
while stack:
node, parent_direction = stack[-1]
if len(directed_graph[node]) == 0:
circuit.append(node[0])
stack.pop()
else:
neighbor, direction = directed_graph[node].pop()
if (neighbor, 1 - direction) != (node[0], parent_direction):
stack.append(((neighbor, direction), parent_direction))
# Combine directed edges that belong to the same undirected edge.
eulerian_circuit = []
for node1, node2 in zip(circuit[:-1], circuit[1:]):
if isinstance(node1, tuple) and isinstance(node2, tuple) and node1[0] == node2[0]:
if node1[1] == 0 and node2[1] == 1:
eulerian_circuit.append(node1[0])
elif node1[1] == 1 and node2[1] == 0:
eulerian_circuit.append(node2[0])
else:
eulerian_circuit.append(node1)
eulerian_circuit.append(circuit[-1][0])
return eulerian_circuit[:-1]
以上为Python的示例代码,需注意一下几点:
参考资料: