Two passes, one theorem

Pass 1: DFS over G, push each node when its DFS finishes. Pass 2: DFS over the REVERSE graph G^T, starting from unvisited nodes in decreasing finish order; each pass-2 tree is exactly one SCC.

Why it is correct: the node u with the largest finish time lies in a "source" SCC C of the condensation. Reversing the graph makes C a sink — no edges leave C in G^T, so its pass-2 DFS cannot escape. Induction on the condensation finishes the proof.

Implementation must be ITERATIVE on both passes; build G^T once.