Editorial : Oriented Journey

Summary

This editorial follows model_sol.cpp: encode the bits of $s-1$ on vertices $2,\ldots,n$ with one bit per vertex, then orient each tree edge so that a BFS from vertex $1$ recovers those bits using xor along edges.

Encoding (Player A, $q=1$)

Let $\mathrm{val} = s - 1$. For each vertex $i \in \{2,\ldots,n\}$, define

$$ X_i = \text{bit }(i-2)\text{ of }\mathrm{val}. $$

Set $X_1 = 0$ as reference.

For each revealed undirected edge $(u,v)$ with $u < v$:

If $X_u = X_v$, output the direction $(u,v)$.
Otherwise output $(v,u)$.

This matches the solveA routine.

Decoding (Player B, $q=2$)

Build the adjacency list of the directed tree. For each edge, map the actual ordered pair $(a,b)$ to an xor “weight” on the undirected edge: $0$ if the edge goes from smaller to larger index, $1$ otherwise (see w in the code).

Root the tree at vertex $1$, set $X_1 = 0$, and BFS: for an edge from curr to nxt with weight $w$, set $X_{\mathrm{nxt}} = X_{\mathrm{curr}} \oplus w$.

Reconstruct

$$ s = 1 + \sum_{i=2}^{n} [X_i = 1] \cdot 2^{i-2}. $$

Why it works

The underlying undirected tree is connected; fixing $X_1$ and xor along edges uniquely determines all $X_i$. The orientation rule makes the xor along each edge equal to $|X_u - X_v|$ in $\{0,1\}$, so the recovered bits match the encoder’s $X$.

Implementation notes

Handle t and q in main, branch to solveA / solveB. Flush output; exit on -1.