i want to implement kruskal's algorithm in python how can i go about representing the tree/graph and what approach should i follow to detect the cycles ?
我想在python中实现kruskal算法我如何去表示树/图以及我应该采用什么方法来检测循环?
2 个解决方案
#1
11
The simplest way of representing it (in my opinion) is by using a dict of arrays lists:
表示它的最简单的方法(在我看来)是使用数组列表的命令:
graph = {}
graph[node_id] = [other_node_id for other_node_id in neighbors(node_id)]
A simple way of finding cycles is by using a BF or DF search:
找到循环的一种简单方法是使用BF或DF搜索:
def df(node):
if visited(node):
pass # found a cycle here, do something with it
visit(node)
[df(node_id) for node_id in graph[node]]
Disclaimer: this is actually a sketch; neighbors(), visited() and visit() are just dummies to represent how the algorithm should look like.
免责声明:这实际上是一个草图;邻居()、visit()和visit()只是表示该算法应该是什么样子的假人。
#2
4
Python Graph API is a good place to start.
Python图形API是一个很好的起点。
For example, NetworkX has a Kruskal's algorithm implementation to find the minimum spanning tree.
例如，NetworkX有一个Kruskal的算法实现来找到最小生成树。
If you want to re-invent the wheel and do it yourself, that is also possible.
如果你想重新发明*并自己动手，那也是可能的。
#1
11
The simplest way of representing it (in my opinion) is by using a dict of arrays lists:
表示它的最简单的方法(在我看来)是使用数组列表的命令:
graph = {}
graph[node_id] = [other_node_id for other_node_id in neighbors(node_id)]
A simple way of finding cycles is by using a BF or DF search:
找到循环的一种简单方法是使用BF或DF搜索:
def df(node):
if visited(node):
pass # found a cycle here, do something with it
visit(node)
[df(node_id) for node_id in graph[node]]
Disclaimer: this is actually a sketch; neighbors(), visited() and visit() are just dummies to represent how the algorithm should look like.
免责声明:这实际上是一个草图;邻居()、visit()和visit()只是表示该算法应该是什么样子的假人。
#2
4
Python Graph API is a good place to start.
Python图形API是一个很好的起点。
For example, NetworkX has a Kruskal's algorithm implementation to find the minimum spanning tree.
例如，NetworkX有一个Kruskal的算法实现来找到最小生成树。
If you want to re-invent the wheel and do it yourself, that is also possible.
如果你想重新发明*并自己动手，那也是可能的。