CodeForces 740D Alyona and a tree 题解:DFS+二分,以及 vector 大法好

题目链接
vector 真的好用~

Problem

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 a_i. 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) ≤ a_u.

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 a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the integers written in the vertices.

The next (n - 1) lines contain two integers each. The i-th of these lines contains integers p_i and w_i (1 ≤ p_i ≤ n, 1 ≤ w_i ≤ 10^9) — the parent of the (i + 1)-th vertex in the tree and the number written on the edge between p_i 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 #1

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

Output #1

1 0 1 0 0

Input #2

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

Output #2

4 3 2 1 0

Notes

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 1 controls the vertex 5).

Solution

水题,很容易想到 \Theta (N^2) 的想法,即两两枚举,显然超时。
考虑枚举一个点,去找控制它的点,从它的父亲开始找,一直往上……我们要找的是 dist(i,j) ≤ a_i 的点。显然 j 不断往上,dist(i,j) 单调递增。所以用二分或者树上倍增找就可以了。

Code

#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
using namespace std;
const int maxn=200005,maxe=400005;
int n,tot=0,a[maxn],fa[maxn],sum[maxn],lnk[maxn],son[maxe],nxt[maxe],w[maxe];
long long dst[maxn];
bool vis[maxn];
vector <int> vec;
inline int read(){
    int ret=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9') ret=ret*10+ch-'0',ch=getchar();
    return ret*f;
}
inline void add(int x,int y,int z){
    tot++;son[tot]=y;w[tot]=z;nxt[tot]=lnk[x];lnk[x]=tot;
}
inline long long myabs(long long x){
    return x>0?x:-x;
}
inline void BuildDistance(int x){
    vis[x]=1;
    for (int i=lnk[x];i;i=nxt[i]) if (!vis[son[i]]){
        fa[son[i]]=x;
        dst[son[i]]=(long long)dst[x]+w[i];
        BuildDistance(son[i]);
    }
}
inline long long GetDistance(int x,int y){
    return myabs((long long)dst[x]-dst[y]);
}
inline void BuildSum(int x){
    vis[x]=1;
    int L=0,R=vec.size()-1,mid,now=-1;
    while (L<=R){
        mid=((R-L)>>1)+L;
        if ((long long)GetDistance(x,vec[mid])<=(long long)a[x]){
            now=mid;
            R=mid-1;
        } else L=mid+1;
    }
    if (now!=-1){
        now=vec[now];
        if (now!=x||GetDistance(x,now)<=a[x]) sum[fa[x]]++,sum[fa[now]]--;
    }
    vec.push_back(x);
    for (int i=lnk[x];i;i=nxt[i]) if (!vis[son[i]]){
        BuildSum(son[i]);
    }
    vec.erase(vec.end()-1);
}
inline void GetAnswer(int x){
    vis[x]=1;
    for (int i=lnk[x];i;i=nxt[i]) if (!vis[son[i]]){
        GetAnswer(son[i]);
        sum[x]+=sum[son[i]];
    }
}
int main(){
    n=read();
    for (int i=1;i<=n;i++) a[i]=read();
    for (int i=2;i<=n;i++){
        int x=read(),y=read();
        add(i,x,y);add(x,i,y);
    }
    BuildDistance(1);
    memset(vis,0,sizeof(vis));
    vec.clear();
    vec.push_back(1);
    BuildSum(1);
    memset(vis,0,sizeof(vis));
    GetAnswer(1);
    for (int i=1;i<=n;i++) printf("%d ",sum[i]);
    printf("\n");
    return 0;
}

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