[做题记录-数据结构] [Ynoi2008] rdCcot

考虑对于每个连通块维护代表元。但是同一深度会有多个代表元， 我们取\(bfs\)序靠前的一个维护。
那么现在给出的结论是一个点是代表元当且仅当它与\(bfs\)序比它小的点没有连边。
那么现在就是要求一个\(pre_i\)表示最大的\(i\)使得\(i < j\)且\(b_i < b_j\)且\(dis(i, j) \leq C\)。第三个限制可以点分治， 然后平衡树上二分得到区间。\(b_i < b_j\)可以把点排序以后按顺序插入平衡树， \(i < j\)也可以在二分之前直接从平衡树上\(split\)下来。
注意点分治的时候\(findrt\)要进行两次！！！

#include <bits/stdc++.h>
#include <bits/extc++.h>

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;

class Input {
	#define MX 1000000
	private :
		char buf[MX], *p1 = buf, *p2 = buf;
		inline char gc() {
			if(p1 == p2) p2 = (p1 = buf) + fread(buf, 1, MX, stdin);
			return p1 == p2 ? EOF : *(p1 ++);
		}
	public :
		Input() {
			#ifdef Open_File
				freopen("a.in", "r", stdin);
				freopen("a.out", "w", stdout);
			#endif
		}
		template <typename T>
		inline Input& operator >>(T &x) {
			x = 0; int f = 1; char a = gc();
			for(; ! isdigit(a); a = gc()) if(a == '-') f = -1;
			for(; isdigit(a); a = gc()) 
				x = x * 10 + a - '0';
			x *= f;
			return *this;
		}
		inline Input& operator >>(char &ch) {
			while(1) {
				ch = gc();
				if(ch != '\n' && ch != ' ') return *this;
			}
		}
		inline Input& operator >>(char *s) {
			int p = 0;
			while(1) {
				s[p] = gc();
				if(s[p] == '\n' || s[p] == ' ' || s[p] == EOF) break;
				p ++; 
			}
			s[p] = '\0';
			return *this;
		}
	#undef MX
} Fin;

class Output {
	#define MX 1000000
	private :
		char ouf[MX], *p1 = ouf, *p2 = ouf;
		char Of[105], *o1 = Of, *o2 = Of;
		void flush() { fwrite(ouf, 1, p2 - p1, stdout); p2 = p1; }
		inline void pc(char ch) {
			* (p2 ++) = ch;
			if(p2 == p1 + MX) flush();
		}
	public :
		template <typename T> 
		inline Output& operator << (T n) {
			if(n < 0) pc('-'), n = -n;
			if(n == 0) pc('0');
			while(n) *(o1 ++) = (n % 10) ^ 48, n /= 10;
			while(o1 != o2) pc(* (--o1));
			return *this; 
		}
		inline Output & operator << (char ch) {
			pc(ch); return *this; 
		}
		inline Output & operator <<(const char *ch) {
			const char *p = ch;
			while( *p != '\0' ) pc(* p ++);
			return * this;
		}
		~Output() { flush(); } 
	#undef MX
} Fout;

#define cin Fin
#define cout Fout
#define endl '\n'

using LL = long long;

inline int log2(unsigned int x);
inline int popcount(unsigned x);
inline int popcount(unsigned long long x);

const int N = 3e5 + 10;
const int INF = 0x3f3f3f3f;

int n, m, C;
struct edge {
	int to, next;
} e[N << 1];
int cnt, head[N];
void push(int x, int y) { e[++ cnt] = (edge) {y, head[x]}; head[x] = cnt; }

static const long long kMul = 0x9ddfea08eb382d69ULL;
long long Seed = 1145141919810;
inline long long fingerPrint(long long x) {
	return x *= kMul, x ^= x >> 47, x *= kMul, x ^= x >> 47, x *= kMul, x ^= x >> 47, x * kMul;
}
inline int frand(void) { return Seed += fingerPrint(Seed); }

struct Node {
	Node *ls, *rs;
	int val, rnd, mndis, dis;
	Node() {}
	Node(int _val, int _dis) : val(_val), rnd(frand()), ls(NULL), rs(NULL), mndis(_dis), dis(_dis) {}
	void upd() {
		mndis = dis;
		if(ls) mndis = min(mndis, ls -> mndis); if(rs) mndis = min(mndis, rs -> mndis);
	}
	void print() {
		if(ls) ls -> print();
		cerr << val << ' ';
		if(rs) rs -> print();
	}
} ;

Node pool[1000000];
Node *oo = pool;

Node *NewNode(int val, int d) {
	oo ++;
	* oo = Node(val, d);
	return oo;
}

class FhqTreap {
	private :
		Node *root;
		Node *merge(Node *x, Node *y) {
			if(x == NULL) return y;
			if(y == NULL) return x;
			if(x -> rnd <= y -> rnd) {
				x -> rs = merge(x -> rs, y);
				x -> upd(); return x;
			}
			else {
				y -> ls = merge(x, y -> ls);
				y -> upd(); return y;
			}
		}
		void split(Node *now, int k, Node *&x, Node *&y) {
			if(now == NULL) { x = NULL; y = NULL; return ; }
			if(now -> val <= k) {
				x = now;
				split(now -> rs, k, x -> rs, y);
			}
			else {
				y = now;
				split(now -> ls, k, x, y -> ls);
			}
			now -> upd();
		}
	public :
		FhqTreap() : root(NULL) { oo = pool; } 
		void Clear() {
			root = NULL; oo = pool;
		}
		void Insert(int k, int dis) {
			Node *a = NULL, *b = NULL, *c = NULL;
			split(root, k, a, b);
			c = NewNode(k, dis);
			root = merge(a, merge(c, b));
		}
		pair<int, int> qry(int i, int d) {
			d = C - d;
			Node *x = NULL, *y = NULL, *a = NULL, *b = NULL;
			split(root, i, x, y); a = x; b = y; 
			int res1 = 0;
			if(x != NULL && x -> mndis <= d) {
				while(x != NULL) {
					if(x -> rs != NULL) {
						if(x -> rs -> mndis <= d) {
							x = x -> rs;
							continue;
						} 
					}
					if(x -> dis <= d) {
						res1 = x -> val; break;
					}
					x = x -> ls;
				}
			}
			int res2 = n + 1;
			if(y != NULL && y -> mndis <= d) {
				while(y != NULL) {
					if(y -> ls != NULL) {
						if(y -> ls -> mndis <= d) {
							y = y -> ls;
							continue;
						} 
					}
					if(y -> dis <= d) {
						res2 = y -> val; break;
					}
					y = y -> rs;
				}
			}
			root = merge(a, merge(NewNode(i, C - d), b));
			return make_pair(res1, res2);
		}
		void print() {
			if(root != NULL) root -> print();
		}
} Treap;

struct Bit {
	int c[N];
	#define lowbit(x) (x & -x)
	void upd(int x, int y) {
		for(; x <= n; x += lowbit(x)) c[x] += y;
	}
	void add(int l, int r, int v) {
		upd(l, v); upd(r + 1, -v);
	}
	int ask(int x) {
		int ans = 0;
		for(; x; x -= lowbit(x)) ans += c[x];
		return ans;
	}
} bit;

int rt, sz[N], all, vis[N];

void findrt(int x, int fx) {
	sz[x] = 1;
	int mx = 0;
	for(int i = head[x]; i; i = e[i].next) {
		int y = e[i].to;
		if(y != fx && ! vis[y]) {
			findrt(y, x);
			sz[x] += sz[y];
			mx = max(mx, sz[y]);
		}
	}
	mx = max(mx, all - sz[x]);
	if(mx * 2 <= all || rt == 0) rt = x;
}

int pre[N], suf[N], bfn[N], fa[N], dist[N];

void getnode(int x, int fx, int nrt, int *tac, int dis) {
	dist[x] = dis;
	if(dis <= C) tac[++ tac[0]] = x;
	else return ;
	for(int i = head[x]; i; i = e[i].next) {
		int y = e[i].to;
		if(! vis[y] && y != fx) getnode(y, x, nrt, tac, dis + 1);
	}
}

void calc(int x) {
	static int tac[N];
	tac[0] = 0;
	getnode(x, 0, x, tac, 0);
	sort(tac + 1, tac + 1 + tac[0], [](int a, int b) { return bfn[a] < bfn[b]; } );
	Treap.Clear();
	for(int p = 1; p <= tac[0]; p ++) {
		int i = tac[p]; int d = dist[i];
		pair<int, int> Get = Treap.qry(i, d);
		pre[i] = max(pre[i], Get.first);
		suf[i] = min(suf[i], Get.second);
	}
}

void solve(int x) {
	vis[x] = 1;
	calc(x);
	for(int i = head[x]; i; i = e[i].next) {
		int y = e[i].to;
		if(! vis[y]) {
			findrt(y, 0);
			all = sz[y]; rt = 0;
			findrt(y, 0);
			solve(rt);
		}
	}
}

struct Line { int l, r, v; };

int Ans[N];
vector<Line> mdf[N];
vector<pair<int, int> > qry[N];

void addmat(int l1, int r1, int l2, int r2) {
	mdf[l1].push_back( {l2, r2, 1} );
	mdf[r1 + 1].push_back( {l2, r2, -1} );
}

int main() {
	cin >> n >> m >> C;
	for(int i = 2, x; i <= n; i ++) {
		cin >> x;
		push(x, i); push(i, x);
	}
	deque<int> q;
	q.push_back(1);
	while(q.size()) {
		int x = q.front(); q.pop_front();
		bfn[x] = ++ bfn[0];
		for(int i = head[x]; i; i = e[i].next) {
			int y = e[i].to;
			if(y != fa[x]) {
				q.push_back(y);
				fa[y] = x;
			}
		}
	}
	for(int i = 1; i <= n; i ++) pre[i] = 0, suf[i] = n + 1;
	all = n; rt = 0;
	findrt(1, 0);
	findrt(rt, 0);
	solve(rt);
	for(int i = 1; i <= n; i ++) addmat(pre[i] + 1, i, i, suf[i] - 1);
	for(int i = 1, l, r; i <= m; i ++) {
		cin >> l >> r;
		qry[l].push_back({r, i});
	}
	for(int i = 1; i <= n; i ++) {
		for(auto p : mdf[i]) 
			bit.add(p.l, p.r, p.v);
		for(auto p : qry[i]) 
			Ans[p.second] = bit.ask(p.first);
	}
	for(int i = 1; i <= m; i ++) cout << Ans[i] << endl;
	return 0;
}

inline int log2(unsigned int x) { return __builtin_ffs(x); }
inline int popcount(unsigned int x) { return __builtin_popcount(x); }
inline int popcount(unsigned long long x) { return __builtin_popcountl(x); }

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