树的点分治（以下简称“点分治”）算法同后缀数组一样，也是很早就接触但几乎没练过的算法。此专题总结一下过去遇到的若干树的点分治题目。
树的重心
的题面给出了树的重心（centroid(s) of a tree）的定义：
Given a tree with $N$ vertices, in order to define the centroid, some integer value will be assosciated to every vertex. Let's consider the vertex $k$. If we remove the vertex $k$ from the tree (along with its adjacent edges), the remaining graph will have only $N-1$ vertices and may be composed of more than one connected components. Each of these components is (obviously) a tree. The value associated to vertex $k$ is the largest number of vertices contained by some connected component in the remaining graph, after the removal of vertex $k$. All the vertices for which the associated value is minimum are considered centroids.
简言之即：树中一点，删去它后所得若干棵树的节点数的最大值最小。
树的重心通过简单的DP即可求得。树的重心的性质 $\DeclareMathOperator{\size}{size}$
以树的重心为根，则子树的点数不超过 $\frac{N}{2}$ 。
证明：将重心记为 $u$，取 $u$ 的某个儿子 $v$，子树 $v$ 的点数记为 $\size(v)$ 。若 $\size(v) > \frac N2$，则以点 $v$ 为根，所有子树的点数最大值将不超过 $\max\{ \size(v) - 1, N - \size(v)\}< \size(v)$，这意味着 $u$ 并不是树的重心。证毕。
注：上述 $\frac N2$ 是在实数下的除法，不是按某种规则取整之后的。