树的点分治专题
树的重心
SGU #134 Centroid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
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。
