codeforces 246 E. Blood Cousins Return (set+dsu on the tree)

E. Blood Cousins Return
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Polycarpus got hold of a family tree. The found tree describes the family relations of n people, numbered from 1 to n. Every person in this tree has at most one direct ancestor. Also, each person in the tree has a name, the names are not necessarily unique.

We call the man with a number a a 1-ancestor of the man with a number b, if the man with a number a is a direct ancestor of the man with a number b.

We call the man with a number a a k-ancestor (k > 1) of the man with a number b, if the man with a number b has a 1-ancestor, and the man with a number a is a (k - 1)-ancestor of the 1-ancestor of the man with a number b.

In the tree the family ties do not form cycles. In other words there isn't a person who is his own direct or indirect ancestor (that is, who is an x-ancestor of himself, for some xx > 0).

We call a man with a number a the k-son of the man with a number b, if the man with a number b is a k-ancestor of the man with a number a.

Polycarpus is very much interested in how many sons and which sons each person has. He took a piece of paper and wrote m pairs of numbers viki. Help him to learn for each pair viki the number of distinct names among all names of the ki-sons of the man with numbervi.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. Next n lines contain the description of people in the tree. The i-th line contains space-separated string si and integer ri (0 ≤ ri ≤ n), where si is the name of the man with a number i, and ri is either the number of the direct ancestor of the man with a number i or 0, if the man with a number i has no direct ancestor.

The next line contains a single integer m (1 ≤ m ≤ 105) — the number of Polycarpus's records. Next m lines contain space-separated pairs of integers. The i-th line contains integers viki (1 ≤ vi, ki ≤ n).

It is guaranteed that the family relationships do not form cycles. The names of all people are non-empty strings, consisting of no more than 20 lowercase English letters.

Output

Print m whitespace-separated integers — the answers to Polycarpus's records. Print the answers to the records in the order, in which the records occur in the input.

Examples
input
6
pasha 0
gerald 1
gerald 1
valera 2
igor 3
olesya 1
5
1 1
1 2
1 3
3 1
6 1
output
2
2
0
1
0
input
6
valera 0
valera 1
valera 1
gerald 0
valera 4
kolya 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1
output
1
0
0
0
2
0
0


题解:set+dsu on the tree

STL大法好!对于每个深度开一个set记录每种颜色的信息。

#include
#include
#include
#include
#include
#include
#define N 200003
using namespace std;
int n,m,size[N],son[N],mark[N],deep[N],tot,col[N];
int ans[N],point[N],head[N],v[N],nxt[N],next[N],u[N],num[N];
map mp;
set dep[N];
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 z){
	tot++; next[tot]=head[x]; head[x]=tot; u[tot]=y; num[tot]=z;
}
void solve(int x,int fa)
{
	size[x]=1; deep[x]=deep[fa]+1;
	for (int i=point[x];i;i=nxt[i]){
		if (v[i]==fa) continue;
		solve(v[i],x);
		size[x]+=size[v[i]];
		if (size[son[x]]100001) continue;
		ans[num[i]]=dep[t].size();
	}
	if (son[x]) mark[son[x]]=0;
	if (!k) clear(x,fa);
}
int main()
{
	freopen("a.in","r",stdin);
	scanf("%d",&n);
	string s; int root=0; int cnt=0;
	for (int i=1;i<=n;i++) {
		cin>>s; int x; scanf("%d",&x);
		add(x,i);
		if (!mp[s]) mp[s]=++cnt;
		col[i]=mp[s];
	}
	//for (int i=1;i<=n;i++) cout<



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