线段树合并都是蓝题了嘛 我可能和时代脱轨了emm...

直接离散化然后合并就好啦w

生病了真难受QAQ

//Love and Freedom.
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
#define inf 20021225
#define mxn 100010
using namespace std;struct node
{int ls,rs,sum;
}t[mxn*40];
int cnt,rt[mxn],ans[mxn];
struct edge
{int to,lt;
}e[mxn];
int val[mxn],in[mxn],ecnt;
int p[mxn],dd[mxn],n;void add(int x,int y)
{e[++ecnt].to=y;e[ecnt].lt=in[x];in[x]=ecnt;
}void pushup(int x)
{t[x].sum = t[t[x].ls].sum+t[t[x].rs].sum;
}int merge(int x,int y,int l,int r)
{if(!x||!y) return x+y;if(l==r){t[x].sum+=t[y].sum;return x;}int mid = (l+r)>>1;t[x].ls=merge(t[x].ls,t[y].ls,l,mid);t[x].rs=merge(t[x].rs,t[y].rs,mid+1,r);pushup(x);  return x;
}void insert(int &x,int l,int r,int d)
{if(!x) x=++cnt; t[x].sum++;if(l==r) return;int mid=(l+r)>>1;if(d<=mid)  insert(t[x].ls,l,mid,d);else    insert(t[x].rs,mid+1,r,d);pushup(x);
}int query(int x,int l,int r,int d)
{if(!x) return 0;if(l==r) return t[x].sum;int mid=(l+r)>>1;if(d<=mid) return query(t[x].ls,l,mid,d)+t[t[x].rs].sum;else  return query(t[x].rs,mid+1,r,d);
}
bool cmp(int x,int y)
{return val[x]<val[y];
}
void dfs(int x)
{for(int i=in[x];i;i=e[i].lt)dfs(e[i].to);insert(rt[x],1,n+1,p[x]);// int tmp = ;for(int i=in[x];i;i=e[i].lt)rt[x]=merge(rt[x],rt[e[i].to],1,n+1);ans[x] = query(rt[x],1,n+1,p[x]+1);
}
int main()
{int f;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&val[i]),dd[i]=i;sort(dd+1,dd+n+1,cmp);for(int i=1;i<=n;i++)    p[dd[i]]=i;for(int i=2;i<=n;i++){scanf("%d",&f);add(f,i);}dfs(1);for(int i=1;i<=n;i++) printf("%d\n",ans[i]);return 0;
}