code forces739B 树上差分

最新推荐文章于 2021-11-02 22:54:00 发布

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本文介绍了一种结合树形动态规划与二分搜索的算法，用于解决一类特定问题：给定一棵带权树及各节点上的管辖距离范围，求每个节点能够直接或间接管辖的节点数量。文章详细讲解了算法思路，包括如何通过二分查找确定管辖范围内的最远节点，并利用差分技巧统计管辖节点数。

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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 thatv 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.

Examples

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

output
1 0 1 0 0

input
5
9 7 8 6 5
1 1
2 1
3 1
4 1

output
4 3 2 1 0

Note

In the example test case the vertex 1 controls the vertex 3, the vertex 3 controls the vertex 5 (note that is doesn't mean the vertex 1controls the vertex 5).

给n个点和被管辖的距离范围，还有边权，问每个点能管辖多少个点

树上二分找到最后满足管辖范围的值，然后差分一下就出来了

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;

struct node
{
    int v,w;
};

int sum[N];
int fa[N][21];
typedef long long ll;
ll dep[N];
int u[N];
int du[N];
vector<node> e[N];

void dfs(int x,int y,ll w)
{
    dep[x]=w;
    du[x]=1;
    fa[x][0]=y;
    for(int i=1;fa[fa[x][i-1]][i-1];i++)
        fa[x][i]=fa[fa[x][i-1]][i-1];
    int pos=x;
    for(int i=20;i>=0;i--)
        if(fa[pos][i]&&dep[x]-dep[fa[pos][i]]<=u[x]) pos=fa[pos][i];

    int gg=fa[pos][0];
    if(dep[x]-dep[gg]>u[x])
    du[gg]--;
    for(int i=0;i<e[x].size();i++)
    {
        int vv=e[x][i].v;
        if(vv==y) continue;
        dfs(vv,x,w+e[x][i].w);
        du[x]+=du[vv];
    }
}

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&u[i]);
    memset(fa,0,sizeof(fa));
    for(int i=1;i<n;i++)
    {
        int x,w;
        scanf("%d %d",&x,&w);
        e[x].push_back((node){i+1,w});
    }
    dep[0]=0;
    dfs(1,0,0);
    for(int i=1;i<=n;i++)
        cout<<du[i]-1<<" ";
}