For a set of stations ~S~, define ~f_A(S)~ as the set of all stations such that if the train is placed on them, Arezou can play in a way that the train reaches one of the stations in ~S~ at some point, regardless of Borzou's moves (therefore ~S \subseteq f_A(S)~ by definition). We define ~f_B(S)~ similarly.

Initially, let ~T = S~, and we iteratively add stations to ~T~ such that eventually ~T~ becomes equal to ~f_A(S)~.

While there exists a station ~v~ that satisfies one of the following conditions, we add ~v~ to the set ~T~:

1. ~v~ is owned by Arezou, and there exists an outgoing track from ~v~ that arrives at a station already in ~T~.
2. ~v~ is owned by Borzou, and all of the outgoing tracks from ~v~ arrive at the stations already in ~T~.

Similarly, we can compute ~f_B(S)~. Notice that both ~f_A(S)~ and ~f_B(S)~ can be computed iteratively in time ~\mathcal O(n+m)~.

Now let ~R~ be the set of all the charging stations. By definition, for every station ~v \notin f_A(R)~, Borzou can win the game if the train is initially placed on ~v~.

Therefore, we can solve the problem as follows:

1. If ~f_A(R)~ is the set of all stations, Arezou can win the game for all initial stations.
2. Otherwise:
   1. Let ~X~ be the set of all stations not in ~f_A(R)~.
   2. Borzou can win the game if the initial stations are in ~f_B(X)~.
   3. Remove ~f_B(X)~ from the graph and solve the problem recursively for the remaining stations.

This algorithm runs in time ~\mathcal O(nm)~.