HDU - 6430 Problem E. TeaTree （启发式合并）（2018 Multi-University Training Contest 10 1005）
Problem E. TeaTree
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 28 Accepted Submission(s): 11
Problem Description
Recently, TeaTree acquire new knoledge gcd (Greatest Common Divisor), now she want to test you.
As we know, TeaTree is a tree and her root is node 1, she have n nodes and n-1 edge, for each node i, it has it’s value v[i].
For every two nodes i and j (i is not equal to j), they will tell their Lowest Common Ancestors (LCA) a number : gcd(v[i],v[j]).
For each node, you have to calculate the max number that it heard. some definition:
In graph theory and computer science, the lowest common ancestor (LCA) of two nodes u and v in a tree is the lowest (deepest) node that has both u and v as descendants, where we define each node to be a descendant of itself.
Input
On the first line, there is a positive integer n, which describe the number of nodes.
Next line there are n-1 positive integers f[2] ,f[3], …, f[n], f[i] describe the father of node i on tree.
Next line there are n positive integers v[2] ,v[3], …, v[n], v[i] describe the value of node i.
n<=100000, f[i]
Output
Your output should include n lines, for i-th line, output the max number that node i heard.
For the nodes who heard nothing, output -1.
Sample Input
4 1 1 3 4 1 6 9
Sample Output
2 -1 3 -1
Source
2018 Multi-University Training Contest 10
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题意：有一棵树，每对节点都会告诉他的最近公共祖先他们的GCD，求每一个节点知道的最大的GCD。
解题思路：先预处理，把每一个节点的所有因子预处理出来。然后对树深搜，回溯的时候记录这颗子树的所有因子，然后做启发式合并，小的并入大的。然后更新最大值即可。
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