一、题目回顾
题目链接：zxa and leaf
Problem Description
zxa have an unrooted tree with n nodes, including (n?1) undirected edges, whose nodes are numbered from 1 to n. The degree of each node is defined as the number of the edges connected to it, and each node whose degree is 1 is defined as the leaf node of the tree.
zxa wanna set each node‘s beautiful level, which must be a positive integer. His unrooted tree has m(1≤m≤n) leaf nodes, k(1≤k≤m)
leaf nodes of which have already been setted their beautiful levels, so
that zxa only needs to set the other nodes‘ beautiful levels.
zxa
is interested to know, assuming that the ugly level of each edge is
defined as the absolute difference of the beautiful levels between two
nodes connected by this edge, and the ugly level of the tree is the
maximum of the ugly levels of **all the edges on this tree**, then what
is the minimum possible ugly level of the tree, can you help him?
Input
The first line contains an positive integer T, represents there are T test cases.
For each test case:
The first line contains two positive integers n and k, represent the tree has n nodes, k leaf nodes of which have already been setted their beautiful levels.
The next (n?1) lines, each line contains two distinct positive integers u and v, repersent there is an undirected edge between node u and node v.
The next k lines, each lines contains two positive integers u and w, repersent node u is a leaf node, whose beautiful level is w.
There is a blank between each integer with no other extra space in one line.
It‘s guaranteed that the input edges constitute a tree.
1≤T≤10,2≤n≤5?104,1≤k≤n,1≤u,v≤n,1≤w≤109
Output
For each test case, output in one line a non-negative integer, repersents the minimum possible ugly level of the tree.
Sample Input
2
3 2
1 2
1 3
2 4
3 9
6 2
1 2
1 3
1 4
2 5
2 6
3 6
5 9
Sample Output
3
1
Hint
If you need a larger stack size, please use #pragma comment(linker, "/STACK:102400000,102400000") and submit your solution using C++.
题意：一棵树n个点，其中有一些点已经有权值，现在给剩下的点安排权值，使得树中相邻两点的之差绝对值的最大值最小。
二、解题思路
- 二分+树形dp（也可以二分+dfs）