Weak, Regular, and Strong connectivity in directed graphs There are 3 types of connectivity when talking about a directed graph $G$.
1) weakly connected - replacing all of $G$'s directed edges with undirected edges produces a connected (undirected) graph.
2) connected - contains a directed path from $u$ to $v$ OR a directed path from $v$ to $u$ for every pair of vertices $u$, $v$
3) strongly connected - contains a directed path from $u$ to $v$ AND a directed path from $v$ to $u$ for every pair of vertices $u$, $v$

Efficient algorithm for determining...
weak connectivity - replace $G$'s directed edges with undirected edges and run DFS and see if it reaches every vertex
connectivity - ???
strong connectivity - run DFS on $G$ and see if it reaches every vertex, and then transpose $G$ (reverse each edge in $G$ to yield a new graph $G^T$) and run DFS on $G^T$ and see if it reaches every vertex (since the existence of a path from $u$ to $v$ in $G^T$ implies the existence of a path from $v$ to $u$ in $G$)
What is an efficient algorithm for determining if a directed graph G is connected in this intermediate sense between weak and strong?
EDIT:
I realized "efficient" is pretty vague. So more specifically, I mean that the algorithms I provided for weak connectivity and strong connectivity run in linear time ($O(V + E)$), and the naive algorithm for the intermediate connectivity of running all-pairs path finding and for each pair of vertices $(u, v)$ seeing if a path exists between them in either direction would run in something like $O(V^3)$ or $O(EV\ln{V})$. So I'm wondering if there is a linear time algorithm (or at least something better than the naive approach) to determining this intermediate connectivity, and what that might be.

To check for "regular connected" (semi-connected) directed graph $G$:
1) Form the condensation of $G$ (linear time), giving a directed acyclic graph $H$
2) If $H$ is a (directed) path,  $G$ is "regular connected". (If $H$ has multiple branches, sources or sinks, $G$ is not "regular connected").
