// The problem states it wants the vertex on which the graph is accessible, and if not, return any vertex.
// This can be trivial since you can return the first item in the topological ordering and always be correct.
// However, if we are required to return true or false, then we must perform this funciton.
//
// G: directed acyclic graph
// T: topological ordering of G
getAccessible(G, T):
	lastVisited = stack()
	lookBack = set()
	for v in T:
		repeat = true
		while repeat:
			if v == T[0]:
				lastVisited.push(v)
				repeat = false
			else if lastVisited.empty()
				return v // G is not accessible, I can return T[0] and be correct according to the problem
			else if lastVisited.peek() -> v exists in G:
				lastVisited.push(v)
				repeat = false
			for u in lookBack:
				if u -> v exists in G:
					lastVisited.push(v)
					repeat = false
					break
			if repeat == true: // none of the vertices in lookBack or lastVisited.peek() lead to v
				lookBack.add(lastVisited.pop())
	// if G is accessible, then lastVisited will be non-empty
	return T[0] // again, I can return T[0] and be correct