B. Alyona and a tree

題目連線：

http://codeforces.com/contest/739/problem/B

Description

Alyona has a tree with n vertices. The root of the tree is the vertex 1. In each vertex Alyona wrote an positive integer, in the vertex i she wrote ai. Moreover, the girl wrote a positive integer to every edge of the tree (possibly, different integers on different edges).

Let's define dist(v, u) as the sum of the integers written on the edges of the simple path from v to u.

The vertex v controls the vertex u (v ≠ u) if and only if u is in the subtree of v and dist(v, u) ≤ au.

Alyona wants to settle in some vertex. In order to do this, she wants to know for each vertex v what is the number of vertices u such that v controls u.

Input

The first line contains single integer n (1 ≤ n ≤ 2·105).

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the integers written in the vertices.

The next (n - 1) lines contain two integers each. The i-th of these lines contains integers pi and wi (1 ≤ pi ≤ n, 1 ≤ wi ≤ 109) — the parent of the (i + 1)-th vertex in the tree and the number written on the edge between pi and (i + 1).

It is guaranteed that the given graph is a tree.

Output

Print n integers — the i-th of these numbers should be equal to the number of vertices that the i-th vertex controls.

Sample Input

5
2 5 1 4 6
1 7
1 1
3 5
3 6

Sample Output

1 0 1 0 0

Hint

題意

給你一棵樹，帶有邊權，然後u被v統治的前提是，u在v的子樹裡面，且dis(u,v)<=a[u]

現在問你每個點，統治多少個點。

題解：

考慮每個點x，如果他的祖先p，deep[x]-a[x]<=deep[p]的話，那麼就會被+1。

我們按照dfs的順序去更新的話，顯然就是每次使得一條鏈區間加1。

所以你樹鏈剖分，或者用字首和去維護，就好了。

#include<iostream>
#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>

using namespace std;

int n;

long long a[200005];
long long dep[200005];

vector< pair<int ,long long> >e[200005];
vector< pair<long long,int > >path;

int ans[200005];

void dfs(int u)
{
    long long ts=dep[u]-a[u];
    path.push_back(make_pair(dep[u],u));
    int id=lower_bound(path.begin(),path.end(),make_pair(ts,-1))-path.begin()-1;
    if(id>=0)
    {
        ans[path[id].second]--;
    }

    for(int i=0;i<e[u].size();i++)
    {
        pair<int ,long long> v=e[u][i];
        dep[v.first]=dep[u]+v.second;
        dfs(v.first);
        ans[u]+=(1+ans[v.first]);
    }
    path.pop_back();
}

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=0;i<=200000;i++)
        {
            e[i].clear();
            dep[i]=0;
            ans[i]=0;
        }


        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
        }

        for(int i=1;i<=(n-1);i++)
        {
            int p;long long w;
            scanf("%d%lld",&p,&w);
            e[p].push_back(make_pair(i+1,w));
        }

        dfs(1);

        for(int i=1;i<=n;i++)
        {
            printf("%d ",ans[i]);
        }
        printf("\n");
    }
}