E. Split the Tree
time limit per test 2 seconds
memory limit per test 256 megabytes
inputstandard input
outputstandard output

You are given a rooted tree on n vertices, its root is the vertex number 1. The i-th vertex contains a number $w_i$. Split it into the minimum possible number of vertical paths in such a way that each path contains no more than L vertices and the sum of integers $w_{}$

A vertical path is a sequence of vertices $v_1,v_2,…,v_k$ where $v_i (i≥2)$ is the parent of $v_{i−1}$.

Input
The first line contains three integers n, L, S $(1\le n\le 10^5,1\le L\le 10^5,$

The second line contains n integers $w_1,w_2,…,w_n (1≤w_i≤10^9)$ — the numbers in the vertices of the tree.

The third line contains n−1 integers $p_2,…,p_n (1≤p_i<i)$, where $p_i$ is the parent of the i-th vertex in the tree.

Output
Output one number — the minimum number of vertical paths. If it is impossible to split the tree, output −1.

Examples
input
3 1 3
1 2 3
1 1
output
3
input
3 3 6
1 2 3
1 1
output
2
input
1 1 10000
10001
output
-1

思路：對於一個以k為根結點的子樹來說，k的所有兒子都會延伸上來一條鏈，那麼現在就是從這些鏈中選一條鏈，那麼當然是選一條向上延伸最長的啦。
求出這條鏈向上延伸的最長長度就要用到倍增了。

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e5+10;
const int MOD=1e9+7;
const long long INF=2e18;
typedef long long ll;
vector<int>e[MAX];
ll a[MAX];
ll sum[MAX][21];
int nex[MAX][21];
int cal(pair<int,ll>now,int k,ll L,ll S)
{
    int cnt=0;
    for(int i=20;i>=0;i--)
    {
        if(sum[k][i]+now.second>S)continue;
        if((1<<i)+now.first>L)continue;
        L-=1<<i;
        S-=sum[k][i];
        k=nex[k][i];
        cnt+=1<<i;
    }
    return cnt;
}
ll L,S;
int ans=0;
pair<int,ll>dfs(int k,int fa)
{
    nex[k][0]=fa;
    for(int i=1;i<=20;i++)nex[k][i]=nex[nex[k][i-1]][i-1];
    sum[k][0]=a[k];
    for(int i=1;i<=20;i++)sum[k][i]=min(sum[k][i-1]+sum[nex[k][i-1]][i-1],INF);
    pair<int,ll>QWQ=make_pair(0,0);
    int cnt=0;
    for(int i=0;i<e[k].size();i++)
    {
        pair<int,ll> now=dfs(e[k][i],k);
        if(cal(now,k,L,S)>cnt)
        {
            cnt=cal(now,k,L,S);
            QWQ=now;
        }
    }
    if(QWQ.first==0&&ans!=-1)
    {
        if(QWQ.first+1<=L&&QWQ.second+a[k]<=S)ans++;
        else ans=-1;
    }
    return {QWQ.first+1,QWQ.second+a[k]};
}
int main()
{
    int n;
    cin>>n>>L>>S;
    for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
    for(int i=2;i<=n;i++)
    {
        int x;
        scanf("%d",&x);
        e[x].push_back(i);
    }
    memset(nex,0,sizeof nex);
    memset(sum,0,sizeof sum);
    for(int i=0;i<=20;i++)sum[0][i]=INF;
    dfs(1,0);
    cout<<ans<<endl;
    return 0;
}