CF208E Blood Cousins [dsu on tree 树上启发式合并]

倍增，然后数一下有几个深度为 (dep_u) 的

// powered by c++11
// by Isaunoya
#include <bits/stdc++.h>
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
#define Tp template
using pii = pair<int, int>;
#define fir first
#define sec second
Tp<class T> void cmax(T& x, const T& y) {if (x < y) x = y;} Tp<class T> void cmin(T& x, const T& y) {if (x > y) x = y;}
#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
Tp<class T> void sort(vector<T>& v) { sort(all(v)); } Tp<class T> void reverse(vector<T>& v) { reverse(all(v)); }
Tp<class T> void unique(vector<T>& v) { sort(all(v)), v.erase(unique(all(v)), v.end()); }
const int SZ = 1 << 23 | 233;
struct FILEIN { char qwq[SZ], *S = qwq, *T = qwq, ch;
#ifdef __WIN64
#define GETC getchar
#else
  char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
#endif
  FILEIN& operator>>(char& c) {while (isspace(c = GETC()));return *this;}
  FILEIN& operator>>(string& s) {while (isspace(ch = GETC())); s = ch;while (!isspace(ch = GETC())) s += ch;return *this;}
  Tp<class T> void read(T& x) { bool sign = 0;while ((ch = GETC()) < 48) sign ^= (ch == 45); x = (ch ^ 48);
    while ((ch = GETC()) > 47) x = (x << 1) + (x << 3) + (ch ^ 48); x = sign ? -x : x;
  }FILEIN& operator>>(int& x) { return read(x), *this; } FILEIN& operator>>(ll& x) { return read(x), *this; }
} in;
struct FILEOUT {const static int LIMIT = 1 << 22 ;char quq[SZ], ST[233];int sz, O;
  ~FILEOUT() { flush() ; }void flush() {fwrite(quq, 1, O, stdout); fflush(stdout);O = 0;}
  FILEOUT& operator<<(char c) {return quq[O++] = c, *this;}
  FILEOUT& operator<<(string str) {if (O > LIMIT) flush();for (char c : str) quq[O++] = c;return *this;}
  Tp<class T> void write(T x) {if (O > LIMIT) flush();if (x < 0) {quq[O++] = 45;x = -x;}
		do {ST[++sz] = x % 10 ^ 48;x /= 10;} while (x);while (sz) quq[O++] = ST[sz--];
  }FILEOUT& operator<<(int x) { return write(x), *this; } FILEOUT& operator<<(ll x) { return write(x), *this; }
} out;
#define int long long
int n , q ;
const int maxn = 1e5 + 51 ;
vector < int > g[maxn] ;
int fa[maxn] , ans[maxn] ;
int f[maxn][22] ;
vector < pair < int , int > > qr[maxn] ;
int sz[maxn] , son[maxn] , dep[maxn] ;
void dfs(int u) {
	sz[u] = 1 ;
	for(int v : g[u]) {
		f[v][0] = u , dep[v] = dep[u] + 1 ;
		dfs(v) ;
		sz[u] += sz[v] ;
		if(sz[v] > sz[son[u]])
			son[u] = v ;
	}
}

int vis[maxn] , sum[maxn] ;

void upd(int dep , int qwq) {
	sum[dep] += qwq ;
}
void add(int u , int qwq) {
	upd(dep[u] , qwq) ;
	for(int v : g[u]) 
		if(! vis[v] && v ^ fa[u])
			add(v , qwq) ;
}
void dfs(int u , int kep) {
	for(int v : g[u])
		if(v ^ son[u])
			dfs(v , 0) ;
	if(son[u]) {
		dfs(son[u] , 1) ;
		vis[son[u]] = 1 ;
	}
	add(u , 1) ;
	vis[son[u]] = 0 ;
	for(auto x : qr[u])
		ans[x.second] = sum[x.first] - 1 ;
	if(!kep) add(u , -1) ;
}

int kth(int x , int k) {
	for(int i = 20 ; ~ i ; i --)
		if(k & (1 << i))
			x = f[x][i] ;
	return x ;
}
signed main() {
  // code begin.
	in >> n ;
	rep(i , 1 , n) {
		in >> fa[i] ;
		g[fa[i]].pb(i) ;
	}
	rep(i , 1 , n) {
		if(! fa[i])
			dfs(i) ;
	}
	rep(j , 1 , 20)
		rep(i , 1 , n) 
			f[i][j] = f[f[i][j - 1]][j - 1] ;
	in >> q ;
	rep(i , 1 , q) {
		int v , p ;
		in >> v >> p ;
		int u = kth(v , p) ;
		qr[u].pb(dep[v] , i) ;
	}
	rep(i , 1 , n)
		if(! fa[i]) dfs(i , 0) ;
	rep(i , 1 , q) 
		out << ans[i] << ' ' ;
	out << '
' ;
	return 0;
  // code end.
}