Whenever we have a cycle in our graph, we can direct the edges along the cycle. After
such a change, the reachability will be the same as in the undirected case.
More generally, whenever we have a non-trivial biconnected component, we can direct
its edges to make a strongly connected component in the directed graph. 
(Note that we don't actually have to do that in our code.)

Find all biconnected components and contract each of them to a vertex. You will be
left with a tree. In this tree, we have up to 20 pairs (s[i],t[i]) for which we 
want to guarantee connectivity.

Imagine a graph with one vertex for each (s[i],t[i]) pair. In this graph, connect
two vertices by an edge iff the corresponding requests cannot be both satisfied
(i.e., if they both contain the same edge of the tree and want to use it in opposite
direction).

The answer is the size of the maximum independent set in that graph. As there are
at most 20 vertices, we can find that using brute force.