Nowcoder problem (Chinese) link
First I would like to thank ShaoNianTongXue5307 for his idea!
This is a learning note, most for myself. Most of this blog is not original.
Part 1: Problem Statement:
A tree $$$T=(V, E)$$$ has $$$n$$$ vertices and $$$n-1$$$ edges, the weight of each vertex $$$i$$$ is $$$a_i$$$.
For each edge $$$e$$$, you can determine its direction, i.e., for two vertices $$$u, v$$$, there are two states: $$$u \rightarrow v$$$ and $$$v \rightarrow u$$$. There are $$$2^{n-1}$$$ states in total.
For each state $$$S$$$, we define $$$f(S)$$$ as
$$$f(S) := \sum\limits_{(u, v) \in V \times V, v\,\text{is reachable from}\,u} |a_u - a_v|$$$.
Compute the sum of $$$f(S)$$$ over all $$$2^{n-1}$$$ states $$$S$$$, modulo $$$998244353$$$.
Example:
Suppose $$$n=3$$$, and two edges connect $$$(1, 2), (2, 3)$$$, respectively. $$$a_1 = 3, a_2 = 1, a_3 = 2$$$, the answer is $$$14$$$.
Constraints: $$$2 \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq a_i \leq 10^9$$$.
Part 2: What is the centroid decomposition good at?
Before we learn centroid decomposition, we should first know what it is good at. For a tree $$$T = (V, E)$$$, we can decompose $$$V$$$ into $$$V = V_1 \cup V_2 \cup ... \cup V_t$$$, where $$$V_i (1 \leq i \leq t)$$$ are pairwise disjoint non-empty sets. $$$V_1$$$ is a singleton which contains only one element: The centroid. $$$V_2, ..., V_t$$$ are the connected components of $$$T \setminus V_1$$$. We want to calculate some function $$$f(T)$$$, and we assume that the base case, where $$$V(T) = 1$$$, is easy to calculate. This assumption is not strict, because if we can't even deal with the case where $$$V(T)=1$$$, we can actually achieve nothing. The another important function besides $$$f$$$ is the merge function: $$$merge(V_1, V_2, ..., V_t)$$$. Formally
$$$f(V) = \sum\limits_{i=1}^t f(V_i) + merge(V_1, V_2, ..., V_t)$$$. (1)
Like the merge sort, the number of iterations in the centroid decomposition algorithm is $$$O(log |V|)$$$, so it takes roughly $$$O(log |V| \cdot merge)$$$ times. Therefore, the most important advantage to use centroid decomposition is that the merge function could be computed fast enough. Otherwise, it is even slower than the brute force! In our solution merge is $$$O(|V|log|V|)$$$ so its complexity is $$$O(|V|log^2|V|)$$$.
Part 3: What is a centroid, and what is a centroid decomposition?
For a tree, the vertex $$$v$$$ is called a centroid if and only if for any subtree rooted at v’s child has a size at most half the size of the original tree rooted at $$$v$$$. For example, the centroids for the path A--B--C--D are B and C.
Property $$$1$$$: A tree has one or two centroids.
Proof: First, we prove that one tree has at least one centroid. The centroid could be found using a constructive algorithm. First we specify an original root $$$r$$$. Initialize $$$v$$$ as $$$r$$$. Check whether $$$v$$$ is a centroid. If yes, we have already done. If not, replace $$$v$$$ with $$$v$$$'s heavy child $$$u$$$, i.e., $$$argmax(u)\{size[u] \mid u\,\text{is}\,v\text{'s child}\}$$$. The iteration will terminate since the size if finite. When it terminates, the size of subtrees rooted at $$$v$$$'s children $$$\leq \frac{|V|}{2}$$$. We need to be careful that, when the tree is rooted at $$$v$$$ (instead of $$$r$$$), the parent of $$$v$$$ in the $$$r$$$-rooted tree becomes a child of $$$v$$$ in the $$$v$$$-rooted tree, so we need to check the parent of $$$v$$$ in the $$$r$$$-rooted tree. Since the algorithm does not terminate at $$$v$$$'s parent, the size of $$$v \leq \frac{|V|}{2}$$$, therefore, (the size of $$$v$$$'s parent) — (the size of $$$v$$$'s parent) $$$\leq \frac{|V|}{2}$$$, which also satisfies the condition.
Second, we prove that one tree has at most two centroids. Suppose $$$a$$$ and $$$b$$$ are two centroids. Then, there is a subtree of $$$a$$$'s child ($$$A$$$) containing $$$b$$$, and a subtree of $$$b$$$'s child ($$$B$$$) containing $$$a$$$, $$$A \cup B = V, |A|, |B| \leq \frac{|V|}{2}$$$. By the principle of inclusion-exclusion (PIE), $$$A$$$ and $$$B$$$ must be disjoint, therefore there is an edge $$$ab$$$. So there cannot be $$$\geq 3$$$ centroids, and if it contains $$$2$$$ centroids, these two centroids must be adjacent.
Property $$$2$$$: $$$v := argmin(v)\{max\{size[u] \mid u\,\text{is}\,v\text{'s child}\}\}$$$ is a centroid. This can be proved using the fact that every tree has at least one centroid, so the "best" vertex must be a centroid. In English, for a vertex $$$v$$$, lets the score of $$$v$$$ be the maximum size of subtrees rooted at $$$v$$$'s children. The $$$v$$$ with the minimum score is the centroid.
Property $$$3$$$: $$$v := argmin(v)\sum\limits_{u \in V} d(u, v)$$$.
