You are playing with boxes on a $K_{1, n}$-$\textbf{subdivision}$ graph $G:=(V, E)$ whose number of vertices is odd, i.e., $|V| \equiv 1$ (mod $2$) with a given central point $C$ such that $\forall v \in V - \{C\}, deg(v) \leq 2$. Actually, $C$ is the center of the star graph $G$. Your goal is to move $n:= (|V| - 1)/2$ boxes ($b_1-b_n$) from their starting points to their terminal points. For each vertex $v \in V-\{C\}$, it is either the starting point $s_i$ of a box $b_i$ or the terminal point $t_j$ of a box $b_j$. The central point $C$ has $k$ vacant places $S$ ($S \cap V = \emptyset$) that do not collide with any path. Points in $\{C\} \cup S$ are neither starting points nor terminal points.
For each step, you choose a box $i$ and move it to a place $p \in S \cup V$. You will immediately fail this game if there is another box $b_j \neq b_i$ standing on your path, e.g., in the initial state of Case 1, if you try to move $b_1$ from $s_1$ to $t_1$ you will fail because $b_2$ blocks $b_1$ on $s_2$. If you fail to move some box, you also fail this game. You win this game iff you successfully move $\textbf{all}$ boxes from their starting points to their terminal points.
In Case 1, if $k=0$, you can move $b_3$ from $s_3$ to $t_3$, $b_2$ from $s_2$ to $s_3$, $b_1$ from $s_1$ to $t_1$, $b_2$ from the $s_3$ to $t_2$.
In Case 2, you will also need one vacant point $vp$. You can move $b_1$ from $s_1$ to $vp$, $b_3$ from $s_3$ to $t_3$, and $b_2$ from $s_2$ to $t_2$. Then, move $b_1$ from $v_1$ to $t_1$. If $k=0$ you will fail this game.
In Case 3 you need $3$ vacant places $vp_1$, $vp_2$, $vp_3$. Your move sequence could be:
(1) $b_3$: $s_3 \rightarrow vp_1$; (2) $b_1$: $s_1 \rightarrow vp_2$; (3) $b_2$: $s_2 \rightarrow vp_3$; (4) $b_3$: $vp_1 \rightarrow t_3$; (5) $b_1$: $vp_2 \rightarrow t_1$; (6) $b_2$: $vp_3 \rightarrow t_2$.
Please note that in Case 4 you only need $1$ vacant place $vp$. You can use this $vp$ to move $\{b_4, b_5\}$ to $\{t_4, t_5\}$, respectively. Then you have $4$ empty places, which are $C$, $vp$, $s_4$ (because $b_4$ has been moved to $t_4$), $s_5$. These $4$ places are enough for you to win this game.
$\textbf{TL; DR}$: Given the graph $G$ and $k$ vacant places, if there is a strategy that you can win, print one of your winning strategies, otherwise print $-1$.
Below is my idea: I think this problem is NP-hard. You might build a digraph that describes the dependency of boxes. For $k=0$ we can use topological sorting and for $k>0$, I think it is a feedback vertex set problem (https://en.wikipedia.org/wiki/Feedback_vertex_set), since moving one vertex to the vacant place is equivalent to removing a vertex from the digraph and winning the game seems like breaking all the cycles in the digraph. But I cannot do the reduction. Please help me win this game, or prove this problem is NP-hard (then I will try a greedy strategy).
