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
#include
#include
#include
#include
#define N 200003
using namespace std;
int tot,point[N],v[N],nxt[N],next[N],head[N],c[N],u[N],mark[N];
int deep[N],size[N],son[N],fa[N][20],mi[20],ans[N],num[N],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]]>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【树链剖分|codeforces 208 E. Blood Cousins （dsu on the tree）】 
