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酰基有向图的BFS遍历算法

[英]Algorithm for BFS traveral of an acylic directed graph

原文 2009-07-20 21:54:22 2 1 python/ algorithm/ graph/ dependencies/ traversal

I'm looking for an elegant Python program that does a BFS traveral of a DAG: 我正在寻找一个执行DAG的BFS遍历的优雅Python程序：

Node A is connected to B ( A->B ) if A "depends on" B (think of python package Foo "depending upon" Bar: Foo->Bar). 如果A“依赖” B（将python包Foo“依赖” Bar认为：Foo-> Bar），则将节点A连接到B（ A->B ）。

In a graph of about 7000 such nodes, I want to sort all nodes such that for all possible (i, j) where 1>=i<j<=7000 .. depends(Ni, Nj) is False. 在大约7000个这样的节点的图中，我想对所有节点进行排序，使得对于所有可能的(i, j) ，其中1>=i<j<=7000 depends(Ni, Nj)为False。 depends(A, B) = True if and only if A->B or A "depends on" B .. and Nx is the node occuring in x th position in the sorted list. 当且仅当A->B或A“依赖于” B ..并且Nx是出现在排序列表中第x个位置的节点时，depends（A，B）= True。

Note: A node can have multiple parents. 注意：一个节点可以有多个父节点。 Eg: A->C and B->C. 例如：A-> C和B-> C。 Therefore, according to the above sorting rule, A and B must come before C. 因此，根据上述排序规则，A和B必须排在C之前。

1 个解决方案

If I am reading the question correctly, it looks like you want a topological sort . 如果我正确地阅读了该问题，则看起来您想要一种拓扑排序。 The most efficient algorithm (O(V+E)) for doing this was proposed by Tarjan , and a Python implementation can be found here . Tarjan提出了用于执行此操作的最有效算法（O（V + E）），可以在此处找到Python实现。

Off-topic, but it seems as though your package dependency analogy is reversed; 偏离主题，但好像您的程序包依赖类推倒了； I would think that "A depends on B" would imply "B->A", but of course this will not change the structure of the tree, merely reverse it. 我认为“ A取决于B”将意味着“ B-> A”，但是当然这不会改变树的结构，而只是反转它。