What are possible uses for cross, back and forward edges after the DFS visit?
Think of the following problem: 
Given a directed graph $G = (V,E)$, determine whether there is a pair of vertices $u, v$, s.t $u\neq v$ and $\mathit{and}$ there are at least two $\mathbf{simple}$ paths from $u$ to $v$.

Hint: use the edge labels given by the $DFS$.

Solution:


   If and only if there are any "cross" or "forward" arcs in the tree given by a $DFS$ tour on $G$, then a pair of vertices $u, v$ exists s.t are at least two $\mathbf{simple}$ paths from $u$ to $v$.


Proof:


   1. Suppose no "Cross" or "Forward" Arcs exist. Then the tree given by the $DFS$ is a directed tree, with some "Backwards" arcs. A tree by definition has only one path between every pair of vertices, and below it is shown backwards arcs only add new simple paths.


Consider the backward arc $(u,v)$:

 $\bullet$ for all the ancestors of $v$, it only adds non simple paths.

$\bullet $ for all the descendants of $v$ and ancestors of $u$, denoted by $w$ (vertices between $u$ and $v$ on the tree), it adds one simple path from $w$ to any vertex below (including) $v$ and above (excluding) $w$. All these paths are new because the tree is directed, and all other paths are non-simple, because they contain a cycle.

 $\bullet $ for descendants of $u$, it makes no difference.


2. 

 $\bullet$ Suppose a "Cross" arc exists between $(u,v)$. Then $u$ cannot be the root of the tree (the proof is trivial and omitted). Which means the following simple paths exist: $r \rightarrow v$, $r \rightarrow u$, $r \rightarrow u \rightarrow v$

 $\bullet$ Suppose a "Forward" arc exists between $(u,v)$. Then $v$ is a descendant of $u$, but not a child of $u$, therefore there are at least two simple paths from $u$ to $v$.