I Want to Cycle.

The cycle-related information can be handled using strongly connected components.

First, decompose the directed graph into strongly connected components. Any strongly connected component of size at least $2$ contains a cycle. Since the input contains no self-loops, a strongly connected component of size $1$ does not contain a cycle.

If each strongly connected component is contracted into one vertex, the resulting graph is a DAG. An original vertex can visit a cycle if and only if its strongly connected component can reach a strongly connected component that contains a cycle.

Therefore, we can solve the problem as follows.

• Find all strongly connected components of the graph.
• Mark every strongly connected component of size at least $2$ as a component containing a cycle.
• Build the reversed edges of the strongly connected component DAG.
• Starting from all components containing a cycle, run DFS or BFS along the reversed edges.

Every visited strongly connected component can reach a component containing a cycle in the original direction. Hence, the vertices belonging to unvisited components are exactly the vertices counted in the answer.

The time complexity is $O(N+M)$, and the memory complexity is $O(N+M)$.