Flight Routes Check - Implementation

Implementation:

if numSCCs == 1:
    print "YES"
else:
    print "NO"
    a = -1
    b = -1
    for v from 1 to n:
        if a == -1:
            a = v
        else if scc[v] != scc[a]:
            b = v
            break
    print a, b

Find the first two nodes with different SCC IDs. This guarantees they are in different components.

This runs in $O(V + E)$ time and uses $O(V)$ space.