Read hints first.
Centroid decomposition gives us the advantage of dividing the tree into multiple components, so we can afford to run naive operations within each one — the overall complexity still stays low. Thus, we can alter the logic of the naive solution a little bit and try to find this for each node $$$u$$$ in the component, where $$$c$$$ is the centroid:
How many paths are there in this component that have $$$u$$$ as one endpoint and pass through $$$c$$$?
We can root the component at $$$c$$$ and calculate bitmasks for nodes in a single DFS. From Hint $$$1$$$, we know there are only $$$21$$$ valid bitmasks. Let's call one of them $$$valid$$$. So, for a fixed $$$u$$$, we need a node $$$v$$$ such that $$$mask_u$$$ $$$\oplus$$$ $$$mask_v$$$ $$$=$$$ $$$valid$$$. We can turn this equation into $$$mask_u$$$ $$$\oplus$$$ $$$valid$$$ $$$=$$$ $$$mask_v$$$. But there is one constaint on $$$v$$$ — it should belong to the subtree of a different child of $$$c$$$ than $$$u$$$ belongs to — so the path passes through $$$c$$$. The variety of $$$v$$$ candidates is tight (only $$$21$$$ bitmasks can match with that of $$$u$$$). We can keep count of bitmasks, then add count of $$$mask_u$$$ $$$\oplus$$$ $$$valid$$$ from other subtrees to our answer. Since the path $$$u$$$ — $$$c$$$ — $$$v$$$ also passes through some $$$w$$$ that is an ancestor of $$$u$$$ (rooted at $$$c$$$), we need to update its answer too. We can use propagation from child to parent here — keep the count on $$$u$$$, and when returning to parent inside DFS call, move that count to the parent, and so on. One thing left is the count of paths that have $$$c$$$ as endpoint. We can count them inside the DFS call. That is just the number of nodes with an already valid bitmask.
Overall complexity is $$$O(NlogN)$$$ — each node is traversed $$$logN$$$ times in Decomposition.
