Codeforces Round #358 (Div. 2) C. Alyona and the Tree

Alyona decided to go on a diet and went to the forest to get some apples. There she unexpectedly found a magic rooted tree with root in the vertex
1, every vertex and every edge of which has a number written on.

The girl noticed that some of the tree's vertices are
sad, so she decided to play with them. Let's call vertex
v sad if there is a vertex
u in subtree of vertex
v such that dist(v, u) > au, where
au is the number written on vertex
u, dist(v, u) is the sum of the numbers written on the edges on the path from
v to u.

Leaves of a tree are vertices connected to a single vertex by a single edge, but the root of a tree is a
leaf if and only if the tree consists of a single vertex — root.

Thus Alyona decided to remove some of tree leaves until there will be no any sad vertex left in the tree. What is the minimum number of leaves Alyona needs to remove?

Input
In the first line of the input integer n (1 ≤ n ≤ 105) is given — the number of vertices in the tree.

In the second line the sequence of n integers
a1, a2, ..., an (1 ≤ ai ≤ 109)
is given, where ai is the number written on vertex
i.

The next n - 1 lines describe tree edges:
ith of them consists of two integers
pi and
ci
(1 ≤ pi ≤ n,
 - 109 ≤ ci ≤ 109), meaning that there is an edge connecting vertices
i + 1 and pi with number
ci written on it.

Output
Print the only integer — the minimum number of leaves Alyona needs to remove such that there will be no any sad vertex left in the tree.

Example

Input

9
88 22 83 14 95 91 98 53 11
3 24
7 -8
1 67
1 64
9 65
5 12
6 -80
3 8

Output

5

Note
The following image represents possible process of removing leaves from the tree:



题意：给你一棵树，还有树上节点和边的权值；若一个点的权值小于它到任意一祖先的路径上的权值合，那么以这个点为根的子树就要被删去，问最后剩下多少点。

分析：直接BFS，在BFS的过程中顺便求出每个点和它祖先的最大路径权值和。

#include <cstdio>
#include <queue>
#include <vector>
#include <cstdio>
#include <utility>
#include <cstring>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
long long n,y,val,ans,a[100005],dp[100005],f[100005];
typedef pair<int,long long> pii;
vector <pii> e[100005];
void insert(int x,int y,int val)
{
e[x].push_back(make_pair(y,val));
}
int main()
{
cin.sync_with_stdio(false);
cin>>n;
for(int i = 1;i <= n;i++) cin>>a[i];
for(int i = 1;i < n;i++)
{
cin>>y>>val;
insert(i+1,y,val);
insert(y,i+1,val);
}
deque <int> s;
s.push_front(1);
while(!s.empty())
{
int u = s.front();
for(auto it : e[u])
if(it.first != f[u])
{
f[it.first] = u;
dp[it.first] = max(it.second + dp[u],0ll);
if(dp[it.first] > a[it.first]) continue;
s.push_back(it.first);
ans++;
}
s.pop_front();
}
cout<<n-1-ans<<endl;
}