单连接图? [英] Singly connected Graph?
问题描述
一个单连通图是一个有向图,它最多有 个从u到v∀u,v的路径。
A singly connected graph is a directed graph which has at most 1 path from u to v ∀ u,v.
考虑以下解决方案:
- 从任何顶点运行DFS。
- 现在再次运行DFS,但这一次是从顶点开始,以减少完成时间的顺序。仅对某些先前的DFS中未访问的顶点运行此DFS。如果在同一组件中找到交叉边缘或在前边缘中,则该交叉边缘不是单个连接的。
- 如果所有顶点均已完成且前边缘没有交叉,则将它们单独连接。
O(V + E)
这是对的吗?还是有更好的解决方案。
Is this right? Or is there a better solution.
更新:最多1条简单路径。
Update : atmost 1 simple path.
推荐答案
如果满足以下两个条件之一,则图不会单独连接:
A graph is not singly connected if one of the two following conditions satisfies:
-
在同一组件中,当当您执行DFS时,您会得到一条从一条顶点到另一条已经完成搜索的顶点的路径(当标记为BLACK时)
In the same component, when you do the DFS, you get a road from a vertex to another vertex that has already finished it's search (when it is marked BLACK)
当节点指向时到来自另一个组件的> = 2个顶点,如果2个顶点具有连接,则不会单独连接。但这需要您保留深度优先的森林。
When a node points to >=2 vertices from another component, if the 2 vertices have a connection then it is not singly connected. But this would require you to keep a depth-first forest.
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