We will call Bornholm and Gotland $A$ and $B$ respectively. The first step of our algorithm is setting all the teleports on $A$ in receiving mode, and all teleports on $B$ in sending mode.

Why is this solution not good? The only thing that is wrong is that some receiving teleports on $A$ are useless — there are no teleports sending to them. We will maintain this invariant — after each step the solution will be OK except that some receiving teleports on $A$ will be useless.

In each step we choose a receiving teleport on $A$ with no teleports sending to it. If there are no such teleports, we are finished. Let's call the teleport we have chosen $x$. We switch the teleport $x$ to sending mode. Let $y$ be the destination of $x$ — $y$ is a teleport on $B$. If $y$ is in receiving mode, all is fine — the invariant is true. Otherwise we just switch $y$ to receiving mode. We can make another receiving teleport on $A$ useless this way, but the invariant is still true.

At the end there are no useless teleports left, so we have a correct solution.

At each step the number of receiving teleports on $A$ decreases by $1$, so we will make at most $m$ steps.

To be able to make each step in $\mathcal{O}(1)$ time, we introduce an array $inDegree$, in which we store for each teleport on $A$ the number of teleports sending to it. We also keep track of all useless receiving teleports on $A$ — in the sample solution they are kept on a stack. In each step we just take a teleport from the stack, switch it to sending mode, possibly change the mode of its destination teleport and update one entry in the array $inDegree$.

Reading the input and the initial step take $\mathcal{O}(m+n)$ time, therefore time complexity of the whole algorithm is $\mathcal{O}(m+n)$.