You are given a rooted tree on n vertices, its root is the vertex number 1. The i-th vertex contains a number wi. 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 wi on each path does not exceed S. Each vertex should belong to exactly one path.
A vertical path is a sequence of vertices v1,v2,…,vk where vi (i≥2) is the parent of vi−1.
Input
The first line contains three integers n, L, S (1≤n≤105, 1≤L≤105, 1≤S≤1018) — the number of vertices, the maximum number of vertices in one path and the maximum sum in one path.
The second line contains n integers w1,w2,…,wn (1≤wi≤109) — the numbers in the vertices of the tree.
The third line contains n−1 integers p2,…,pn (1≤pi<i), where pi 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
inputCopy
3 1 3
1 2 3
1 1
outputCopy
3
inputCopy
3 3 6
1 2 3
1 1
outputCopy
2
inputCopy
1 1 10000
10001
outputCopy
-1
Note
In the first sample the tree is split into {1}, {2}, {3}.
In the second sample the tree is split into {1, 2}, {3} or {1, 3}, {2}.
In the third sample it is impossible to split the tree.
题意：
给你n，L,S表示n个点，每个点有一个权值，让你分成最少的链，并且这个链上每个点都是祖先关系，每条链最长是L，链上点的权值之和不超过S。最后给你从2开始的所有点的父亲，1是根。
题解：
首先我们可以肯定的是对于每一个点都是从叶子找上去的，否则会很难找到最优解.
dfs1找的是，当前这个点在满足L和S的情况下能找到的深度最小的祖父，top记录的是这个点的最远祖父下标。
dfs2找的就是答案了。若一个点是叶子结点，那么ans必然+1，如果儿子的最优解是儿子本身，那么也+1.在dfs2中，除非sta[ne]==ne，其它时候的sta都是暂定的。举个例子：
5 3 12
1 1 1 10 1
1 2 2 4
在找5的时候，sta[5]=2，sta[4]=2，但是到2的时候他会发现sta[3]=1,是更优解。
5 4 12
1 1 10 1 1
1 2 2 4这种情况的时候，找到2的时候sta[4]就比sta[3]更优了，所以这种做法是找当前点的最优解。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1e5+5;
struct node
{
int to,next;
}e[N];
int head[N],cnt,fa[N][20],top[N],dep[N];
int n,L;
ll S,w[N],sum[N];
void add(int x,int y)
{
e[cnt].to=y;
e[cnt].next=head[x];
head[x]=cnt++;
}
void bfs()
{
fa[1][0]=1;
dep[1]=0;
queue<int>Q;
Q.push(1);
while(!Q.empty())
{
int u,v;
u=Q.front();
Q.pop();
for(int i=1;i<=16;i++)
fa[u][i]=fa[fa[u][i-1]][i-1];
for(int i=head[u];~i;i=e[i].next)
{
v=e[i].to;
if(v==fa[u][0])
continue;
dep[v]=dep[u]+1;
fa[v][0]=u;
Q.push(v);
}
}
}
void dfs1(int x,int f)
{
sum[x]=sum[f]+w[x];
dep[x]=dep[f]+1;
top[x]=x;
int dis=L;
for(int i=16;i>=0;i--)
{
int gf=fa[top[x]][i];
if(!gf)
continue;
if((1<<i)>=dis)
continue;
if(sum[x]-sum[gf]+w[gf]>S)
continue;
top[x]=gf;
dis-=(1<<i);
}
for(int i=head[x];~i;i=e[i].next)
dfs1(e[i].to,x);
}
int ans,sta[N];
void dfs2(int x)
{
int maxn=-1;
for(int i=head[x];~i;i=e[i].next)
{
int ne=e[i].to;
dfs2(ne);
if(sta[ne]==ne)
continue;
if(maxn==-1||dep[sta[ne]]<dep[maxn])
maxn=sta[ne];
}
if(maxn==-1)
maxn=top[x],ans++;
sta[x]=maxn;
}
int main()
{
memset(head,-1,sizeof(head));
scanf("%d%d%lld",&n,&L,&S);
for(int i=1;i<=n;i++)
{
scanf("%lld",&w[i]);
if(w[i]>S)
return 0*printf("-1\n");
}
int x;
for(int i=2;i<=n;i++)
{
scanf("%d",&x);
add(x,i);
}
bfs();
dfs1(1,0);
dfs2(1);
printf("%d\n",ans);
return 0;
}