黑白树

思路：从子节点染色然到父节点，我们看每个节点的覆盖长度还有他子节点的覆盖长度，如果他的覆盖长度dep[u]>dp[v]-1他子节点的覆盖长度那么我们就去用大的更新dep[u]=max(dep[u],dep[v]-1)然后我们记录一下当覆盖距离为0时，他就只能靠自己去覆盖

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
//#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=100000000;
const int N=2e6+10;
const int M=1e3+10;
const int inf=0x7f7f7f7f;
const int maxx=2e5+7;

ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

ll lcm(ll a,ll b)
{
    return a*(b/gcd(a,b));
}

template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-')
            op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op)
        x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0)
        x = -x, putchar('-');
    if(x >= 10)
        write(x / 10);
    putchar('0' + x % 10);
}
ll qsm(int a,int b,int p)
{
    ll res=1%p;
    while(b)
    {
        if(b&1) res=res*a%p;
        a=1ll*a*a%p;
        b>>=1;
    }
    return res;
}
struct node
{
    int to,nex;
}edge[N];
int san[N];
int unsan[N];
int head[N],tot;
int  ans;
void add(int u,int v)
{
    edge[++tot].to=v;
    edge[tot].nex=head[u];
    head[u]=tot;
}
void dfs(int u,int fa)
{

    for(int i=head[u];i!=0;i=edge[i].nex)
    {
        int v=edge[i].to;
        if(v==fa)continue;
        dfs(v,u);
        san[u]=max(san[u],san[v]-1);
        unsan[u]=max(unsan[u],unsan[v]-1);

    }
    if(unsan[u]==0)
    {
        unsan[u]=san[u];ans++;
    }
}
int main()
{
    SIS;
    int n,k;
    cin>>n;
   // memset(head,-1,sizeof head);
    for(int i=2;i<=n;i++)
    {
        int u;
        cin>>u;
        add(u,i);
        //add(i,u);
    }
    for(int i=1;i<=n;i++)
        cin>>san[i];
   dfs(1,-1);

   cout<<ans<<endl;



    return 0;
}