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

C. Alyona and the Tree

题目连接：

http://www.codeforces.com/contest/682/problem/C

Description

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.

Sample 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

Sample Output

5

Hint

题意

给你一棵树，有点权有边权。

如果存在两个点，u,v。满足u存在v的子树中，(u,v)的之间的边权和大于a[u]的话，那么u点是不开心的。

你只能从叶子节点开始删除点，问你最少删除多少个点，可以使得这个树里面没有不开心的点。

题解：

题目中的树是有顺序的，所以直接从1号节点dfs就好了

如果要删除u点的话，那么显然u子树也得全部删去，所以直接dfs一波就完了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+6;
vector<pair<int,int> >E[maxn];
int a[maxn];
int dfs(int x,int fa,long long dis)
{
    if(dis>a[x])return 0;
    long long ans = 1;
    for(int i=0;i<E[x].size();i++)
    {
        if(E[x][i].first==fa)continue;
        ans+=dfs(E[x][i].first,x,max(dis+E[x][i].second,0LL));
    }
    return ans;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    int p,c;
    for(int i=1;i<=n-1;i++)
    {
        scanf("%d%d",&p,&c);
        E[i+1].push_back(make_pair(p,c));
        E[p].push_back(make_pair(i+1,c));
    }
    int ans = dfs(1,0,0);
    cout<<n-ans<<endl;
}