Codeforces 682C C. Alyona and the Tree (DFS)

C. Alyona and the Tree

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

988 22 83 14 95 91 98 53 113 247 -81 671 649 655 126 -803 8

output

5

Note

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

代码：

#include<stdio.h>#include<string.h>#include<vector>using namespace std;const int maxn=1e5+5;vector<pair<int,int> >E[maxn];int a[maxn];int dfs(int x,int fa,int dis){    if(dis>a[x])return 0;    int ans=1;    for(int i=0;i<E[x].size();i++)    {        if(E[x][i].first==fa)continue;        ans=ans+dfs(E[x][i].first,x,max(E[x][i].second+dis),0);    }    return ans;}int main(){    int t;    scanf("%d",&t);    for(int i=1;i<=t;i++)    {        scanf("%d",&a[i]);    }    for(int i=2;i<=t;i++)    {        int x,y;        scanf("%d%d",&x,&y);        E[i].push_back(make_pair(x,y));        E[x].push_back(make_pair(i,y));    }    int ans=dfs(1,0,0);    printf("%d\n",t-ans);}

题目大意：在一棵树上存在这样一个种节点   u 。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。