codeforces 208 E. Blood Cousins （dsu on the tree）

E. Blood Cousins

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Polycarpus got hold of a family relationship tree. The tree describes family relationships of n people, numbered 1 through n.
Each person in the tree has no more than one parent.

Let's call person a a 1-ancestor of person b,
if a is the parent of b.

Let's call person a a k-ancestor (k > 1) of
person b, if person b has
a 1-ancestor, and a is a (k - 1)-ancestor
of b's 1-ancestor.

Family relationships don't form cycles in the found tree. In other words, there is no person who is his own ancestor, directly or indirectly (that is, who is an x-ancestor
for himself, for some x, x > 0).

Let's call two people x and y (x ≠ y) p-th
cousins (p > 0), if there is person z,
who is a p-ancestor of x and
a p-ancestor of y.

Polycarpus wonders how many counsins and what kinds of them everybody has. He took a piece of paper and wrote m pairs of integersvi, pi.
Help him to calculate the number of pi-th
cousins that person vi has,
for each pair vi, pi.

Input

The first input line contains a single integer n (1 ≤ n ≤ 105) —
the number of people in the tree. The next line contains n space-separated integers r1, r2, ..., rn,
where ri (1 ≤ ri ≤ n) is
the number of person i's parent or 0,
if person i has no parent. It is guaranteed that family relationships don't form cycles.

The third line contains a single number m (1 ≤ m ≤ 105) —
the number of family relationship queries Polycarus has. Next m lines contain pairs of space-separated integers. The i-th
line contains numbers vi, pi (1 ≤ vi, pi ≤ n).

Output

Print m space-separated integers — the answers to Polycarpus' queries. Print the answers to the queries in the order, in which the
queries occur in the input.

Examples

input

6
0 1 1 0 4 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1

output

0 0 1 0 0 1 1

题目大意：给出一棵家谱树，定义向上走k步到达的节点为该点的k-ancestor.每次询问与v同P-ancestor的节点有多少个。

题解：dsu on the tree

将问题转换成p-ancestor的子树有多少个深度为deep[v]的节点。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 200003
using namespace std;
int tot,point
,v
,nxt
,next
,head
,c
,u
,mark
;
int deep
,size
,son
,fa
[20],mi[20],ans
,num
,n,m;
void add(int x,int y)
{
tot++; nxt[tot]=point[x]; point[x]=tot; v[tot]=y;
tot++; nxt[tot]=point[y]; point[y]=tot; v[tot]=x;
}
void build(int x,int y,int i)
{
tot++; next[tot]=head[x]; head[x]=tot; u[tot]=y; c[tot]=i;
}
void solve(int x,int f)
{
deep[x]=deep[f]+1; size[x]=1;
for (int i=1;i<=17;i++) {
if (deep[i]-mi[i]<0) continue;
fa[x][i]=fa[fa[x][i-1]][i-1];
}
for (int i=point[x];i;i=nxt[i]){
if(v[i]==f) continue;
fa[v[i]][0]=x;
solve(v[i],x);
size[x]+=size[v[i]];
if (size[son[x]]<size[v[i]]) son[x]=v[i];
}
}
int get(int x,int k)
{
for (int i=0;i<=17;i++)
if ((k>>i)&1) x=fa[x][i];
return x;
}
void change(int x,int f,int val)
{
num[deep[x]]+=val;
for (int i=point[x];i;i=nxt[i])
if (v[i]!=f&&!mark[v[i]]) change(v[i],x,val);
}
void dfs(int x,int f,bool k)
{
for (int i=point[x];i;i=nxt[i])
if (v[i]!=f&&v[i]!=son[x]) dfs(v[i],x,0);
if (son[x]) dfs(son[x],x,1),mark[son[x]]=1;
change(x,f,1);
for (int i=head[x];i;i=next[i])
ans[c[i]]=num[u[i]];
if (son[x]) mark[son[x]]=0;
if (!k) change(x,f,-1);
}
int main()
{
freopen("a.in","r",stdin);
scanf("%d",&n);
for (int i=1;i<=n;i++) {
int x; scanf("%d",&x);
add(x,i);
}
solve(0,0);
scanf("%d",&m); tot=0;
for (int i=1;i<=m;i++) {
int x,p; scanf("%d%d",&x,&p);
if (deep[x]-2<p) {
ans[i]=1;
continue;
}
int f=get(x,p);
build(f,deep[x],i);
}
dfs(0,0,0);
for (int i=1;i<=m;i++) printf("%d ",ans[i]-1);
printf("\n");
}