【LNOI2014】LCA

题目链接：https://www.luogu.com.cn/problem/P4211

题目大意：给定一棵有根树，对于 (q) 个询问 (l, r, z), 求 (sumlimits_{l leq i leq r} {depth(Lca(i, z))})

solution

考虑另一个问题：求出(sumlimits_{i leq n} {depth(Lca(i, z))}) , 此问题可以转化为 (i) 从 1 到 (n), 分别把 1 到 (i) 路径上的所有点权值加 1, 再求 从根 到 (z) 路径上的权值之和

不难发现 , 原问题可以转化为(sumlimits_{i leq r} {depth(Lca(i, z))} - sumlimits_{i leq l - 1} {depth(Lca(i, z))}) , 可以把原问题分解成两个子问题并离线 , 从 1 到 (n) 分别用树剖修改 , 在此过程中同时算出子问题的答案 , 最后统计即可

时间复杂度: (O(nlog^2n))

code

#include<bits/stdc++.h>
using namespace std;
template <typename T> inline void read(T &FF) {
	int RR = 1; FF = 0; char CH = getchar();
	for(; !isdigit(CH); CH = getchar()) if(CH == '-') RR = -RR;
	for(; isdigit(CH); CH = getchar()) FF = FF * 10 + CH - 48;
	FF *= RR;
}
inline void file(string str) {
	freopen((str + ".in").c_str(), "r", stdin);
	freopen((str + ".out").c_str(), "w", stdout);
}
#define mod 201314
const int N = 1e5 + 10;
int now, fst[N], nxt[N], num[N], l[N], r[N], z[N], rs[N], dep[N];
int n, q, si, ni, pi = 1, size[N], son[N], fa[N], tp[N], rev[N], sg[N]; 
void add(int u, int v) {
	nxt[++now] = fst[u], fst[u] = now, num[now] = v;
	nxt[++now] = fst[v], fst[v] = now, num[now] = u;
}
struct Segment_tree{
	int val, tag;
}xds[N << 2];
struct Que{
	int qi, fi, id, fl;
	friend bool operator < (Que ai, Que bi) {
		return ai.qi < bi.qi;
	}
}qy[N];
void push_up(int pos) {
	xds[pos].val = (xds[pos << 1].val + xds[pos << 1 | 1].val) % mod;
}
void Add(int pos, int l, int r, int ki) {
	xds[pos].val = (xds[pos].val + (r - l + 1) * ki % mod) % mod;
	xds[pos].tag = (xds[pos].tag + ki) % mod;
}
void push_down(int pos, int l, int r) {
	if(xds[pos].tag == 0) return;
	int mid = (l + r) >> 1;
	Add(pos << 1, l, mid, xds[pos].tag);
	Add(pos << 1 | 1, mid + 1, r, xds[pos].tag);
	xds[pos].tag = 0;
}
void modify(int pos, int l, int r, int ll, int rr) {
	if(l >= ll && r <= rr) {
		Add(pos, l, r, 1);
		return;
	}
	push_down(pos, l, r); int mid = (l + r) >> 1;
	if(mid >= ll) modify(pos << 1, l, mid, ll, rr);
	if(mid < rr) modify(pos << 1 | 1, mid + 1, r, ll, rr);
	push_up(pos);
} 
int query(int pos, int l, int r, int ll, int rr) {
	if(l >= ll && r <= rr) return xds[pos].val;
	push_down(pos, l, r); int mid = (l + r) >> 1, res = 0;
	if(mid >= ll) res = (res + query(pos << 1, l, mid, ll, rr)) % mod;
	if(mid < rr) res = (res + query(pos << 1 | 1, mid + 1, r, ll, rr)) % mod;
//	push_up(pos);
	return res;
}
void tree_dsu(int xi) {
	size[xi] = 1; dep[xi] = dep[fa[xi]] + 1;
	for(int i = fst[xi]; i; i = nxt[i])
		if(num[i] != fa[xi]) {
			tree_dsu(num[i]);
			size[xi] += size[num[i]];
			if(size[num[i]] > size[son[xi]])
				son[xi] = num[i];
		}
}
void tree_pre(int xi) {
	if(son[xi]) {
		tp[son[xi]] = tp[xi], sg[++si] = son[xi];
		rev[son[xi]] = si, tree_pre(son[xi]);
	}
	for(int i = fst[xi]; i; i = nxt[i])
		if(num[i] != son[xi] && num[i] != fa[xi]) {
			tp[num[i]] = num[i], sg[++si] = num[i];
			rev[num[i]] = si, tree_pre(num[i]);
		}
}
void modify_list(int xi) {
	int yi = 1;
	while(tp[xi] != tp[yi]) {
		if(dep[tp[xi]] < dep[tp[yi]]) swap(xi, yi);
		modify(1, 1, n, rev[tp[xi]], rev[xi]);
		xi = fa[tp[xi]];
	}
	if(dep[xi] > dep[yi]) swap(xi, yi);
	modify(1, 1, n, rev[xi], rev[yi]);
}
int query_list(int xi) {
	int yi = 1, res = 0;
	while(tp[xi] != tp[yi]) {
		if(dep[tp[xi]] < dep[tp[yi]]) swap(xi, yi);
		res = (res + query(1, 1, n, rev[tp[xi]], rev[xi])) % mod;
		xi = fa[tp[xi]];
	}
	if(dep[xi] > dep[yi]) swap(xi, yi);
	res = (res + query(1, 1, n, rev[xi], rev[yi])) % mod;
	return res;
}
int main() {
	//file("");
	read(n), read(q);
	for(int i = 2; i <= n; i++) {
		read(fa[i]); fa[i]++;
		add(fa[i], i);
	}
	tp[1] = rev[1] = sg[1] = si = 1;
	tree_dsu(1), tree_pre(1);
	for(int i = 1; i <= q; i++) {
		read(l[i]), read(r[i]), read(z[i]); l[i]++, r[i]++, z[i]++;
		if(l[i] != 1) qy[++ni].qi = l[i] - 1, qy[ni].id = i, qy[ni].fi = z[i], qy[ni].fl = -1;
		qy[++ni].qi = r[i], qy[ni].id = i, qy[ni].fi = z[i], qy[ni].fl = 1;
	}
	sort(qy + 1, qy + ni + 1);
	for(int i = 1; i <= n; i++) {
		modify_list(i);
	//	cout << query(1, 1, n, sg[2], sg[2]) << endl;
		while(pi <= ni && qy[pi].qi == i)
			rs[qy[pi].id] += qy[pi].fl * query_list(qy[pi].fi), pi++;
	}
	for(int i = 1; i <= q; i++)
		cout << (rs[i] + mod) % mod << endl;
	return 0;
}