This problem is equivalent to Minimal Labels from Codeforces. Treat the "label" of a vertex in "Minimal Labels" as the completion time of a course in "Course Schedule II." So it suffices to solve the CF problem (editorial) and then output the inverse permutation.

Implementation

Time Complexity: $\mathcal{O}(N + \log N)$

Copy
import heapq

n, m = map(int, input().split())
out = [0] * (n + 1)  # Number of outgoing nodes
radj = [[] for _ in range(n + 1)]  # Reverse graph

for _ in range(m):
	a, b = map(int, input().split())
	radj[b].append(a)
	out[a] += 1