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 $\mathcal{O}(nm)$.