#P3605. [USACO17JAN] Promotion Counting P

    ID: 2451 远端评测题 1000ms 128MiB 尝试: 0 已通过: 0 难度: 7 上传者: 标签>2017线段树USACO树状数组离散化深度优先搜索,DFS

[USACO17JAN] Promotion Counting P

Description

The cows have once again tried to form a startup company, failing to remember from past experience that cows make terrible managers!

The cows, conveniently numbered 1N(1N100,000)1 \ldots N (1 \leq N \leq 100,000), organize the company as a tree, with cow 1 as the president (the root of the tree). Each cow except the president has a single manager (its "parent" in the tree). Each cow i has a distinct proficiency rating, p(i)p(i), which describes how good she is at her job. If cow i is an ancestor (e.g., a manager of a manager of a manager) of cow jj, then we say jj is a subordinate of ii.

Unfortunately, the cows find that it is often the case that a manager has less proficiency than several of her subordinates, in which case the manager should consider promoting some of her subordinates. Your task is to help the cows figure out when this is happening. For each cow ii in the company, please count the number of subordinates jj where p(j)>p(i)p(j) \gt p(i).

Input Format

The first line of input contains NN.

The next NN lines of input contain the proficiency ratings p(1)p(N)p(1) \ldots p(N) for the cows. Each is a distinct integer in the range 11,000,000,0001 \ldots 1,000,000,000.

The next N1N-1 lines describe the manager (parent) for cows 2N2 \ldots N. Recall that cow 1 has no manager, being the president.

Output Format

Please print NN lines of output. The ii-th line of output should tell the number of subordinates of cow ii with higher proficiency than cow ii.

5
804289384
846930887
681692778
714636916
957747794
1
1
2
3
2
0
1
0
0