Description

给出一个n个节点的有根树（编号为0到n-1，根节点为0）。一个点的深度定义为这个节点到根的距离+1。
设dep[i]表示点i的深度，LCA(i,j)表示i与j的最近公共祖先。
有q次询问，每次询问给出l r z，求sigma_{l<=i<=r}dep[LCA(i,z)]。
（即，求在[l,r]区间内的每个节点i与z的最近公共祖先的深度之和）

Solution

将问题转化一下：将【BZOJ3626】【LNOI2014】LCA（树链剖分，线段树）_#define的所有点到根的路径全部加一，答案为【BZOJ3626】【LNOI2014】LCA（树链剖分，线段树）_#define_02到根的路径上的权值和。

我们将询问离线，变成【BZOJ3626】【LNOI2014】LCA（树链剖分，线段树）_最近公共祖先_03的答案减去【BZOJ3626】【LNOI2014】LCA（树链剖分，线段树）_Code_04的答案，然后从【BZOJ3626】【LNOI2014】LCA（树链剖分，线段树）_#define_05到【BZOJ3626】【LNOI2014】LCA（树链剖分，线段树）_Code_06扫一遍，依次将其到根的路径加一，然后查询答案即可。

Code

/************************************************
 * Au: Hany01
 * Date: Sep 6th, 2018

 * Inst: Yali High School
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef long double LD;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define SZ(a) ((int)(a).size())
#define ALL(a) a.begin(), a.end()
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
	static int _, __; static char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 5e4 + 5, MOD = 201314;

int n, beg[maxn], e, v[maxn], nex[maxn], fa[maxn], sz[maxn], top[maxn], dep[maxn], id[maxn], dfn[maxn], clk, son[maxn], ans[maxn];

inline int ad(int x, int y) { return (x += y) >= MOD ? x - MOD : x; }
inline void add(int uu, int vv) { v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e; }

struct Qry { int i, u, id, ty; };
bool operator < (Qry A, Qry B) { return A.i < B.i; }

void DFS1(int u) {
	dep[u] = dep[fa[u]] + 1, sz[u] = 1;
	for (register int i = beg[u]; i; i = nex[i])
		if (v[i] != fa[u]) {
			fa[v[i]] = u, DFS1(v[i]), sz[u] += sz[v[i]];
			if (sz[son[u]] < sz[v[i]]) son[u] = v[i];
		}
}

void DFS2(int u) {
	dfn[u] = ++ clk;
	if (son[u]) top[son[u]] = top[u], DFS2(son[u]);
	for (register int i = beg[u]; i; i = nex[i])
		if (v[i] != fa[u] && v[i] != son[u]) top[v[i]] = v[i], DFS2(v[i]);
}

struct SegmentTree {
	int tr[maxn << 2], tag[maxn << 2];
#define lc (t << 1)
#define rc (lc | 1)
#define mid ((l + r) >> 1)
	inline void pushdown(int t, int l, int r) {
		if (tag[t]) tag[lc] = ad(tag[lc], tag[t]), tag[rc] = ad(tag[rc], tag[t]), tr[lc] = ad(tr[lc], (LL)(mid - l + 1) * tag[t] % MOD), tr[rc] = ad(tr[rc], (LL)(r - mid) * tag[t] % MOD), tag[t] = 0;
	}
	void update(int t, int l, int r, int x, int y) {
		if (x <= l && r <= y) { tr[t] = ad(tr[t], (r - l + 1) % MOD), (++ tag[t]) %= MOD; return; }
		pushdown(t, l, r);
		if (x <= mid) update(lc, l, mid, x, y);
		if (y >  mid) update(rc, mid + 1, r, x, y);
		tr[t] = ad(tr[lc], tr[rc]);
	}
	int query(int t, int l, int r, int x, int y) {
		if (x <= l && r <= y) return tr[t];
		pushdown(t, l, r);
		if (y <= mid) return query(lc, l, mid, x, y);
		if (x >  mid) return query(rc, mid + 1, r, x, y);
		return ad(query(lc, l, mid, x, y), query(rc, mid + 1, r, x, y));
	}
}ST;

inline void update(int u) {
	while (top[u] != 1) ST.update(1, 1, n, dfn[top[u]], dfn[u]), u = fa[top[u]];
	ST.update(1, 1, n, 1, dfn[u]);
}

inline int query(int u) {
	int ans = 0;
	while (top[u] != 1) ans = ad(ST.query(1, 1, n, dfn[top[u]], dfn[u]), ans), u = fa[top[u]];
	ans = ad(ans, ST.query(1, 1, n, 1, dfn[u]));
	return ans;
}

int main()
{
#ifdef hany01
	freopen("bzoj3626.in", "r", stdin);
	freopen("bzoj3626.out", "w", stdout);
#endif

	static int m, l, r, z, qs, cur = 1;
	static Qry Q[maxn << 1];

	n = read(), m = read();
	For(i, 2, n) add(fa[i] = read() + 1, i);
	DFS1(1), DFS2(top[1] = 1);

	For(i, 1, m) {
		l = read() + 1, r = read() + 1, z = read() + 1;
		Q[++ qs] = (Qry){r, z, i, 1};
		if (l > 1) Q[++ qs] = (Qry){l - 1, z, i, 0};
	}
	sort(Q + 1, Q + 1 + qs);

	For(i, 1, n) for (update(i); Q[cur].i == i; ++ cur)
		ans[Q[cur].id] = ad(ans[Q[cur].id], Q[cur].ty ? query(Q[cur].u) : MOD - query(Q[cur].u));
	For(i, 1, m) printf("%d\n", ans[i]);

	return 0;
}

/************************************************
 * Au: Hany01
 * Date: Sep 6th, 2018
 * Prob: BZOJ3626 LNOI2014 LCA
 * Email: hany01dxx@gmail.com & hany01@foxmail.com
 * Inst: Yali High School
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef long double LD;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define SZ(a) ((int)(a).size())
#define ALL(a) a.begin(), a.end()
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
	static int _, __; static char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 5e4 + 5, MOD = 201314;

int n, beg[maxn], e, v[maxn], nex[maxn], fa[maxn], sz[maxn], top[maxn], dep[maxn], id[maxn], dfn[maxn], clk, son[maxn], ans[maxn];

inline int ad(int x, int y) { return (x += y) >= MOD ? x - MOD : x; }
inline void add(int uu, int vv) { v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e; }

struct Qry { int i, u, id, ty; };
bool operator < (Qry A, Qry B) { return A.i < B.i; }

void DFS1(int u) {
	dep[u] = dep[fa[u]] + 1, sz[u] = 1;
	for (register int i = beg[u]; i; i = nex[i])
		if (v[i] != fa[u]) {
			fa[v[i]] = u, DFS1(v[i]), sz[u] += sz[v[i]];
			if (sz[son[u]] < sz[v[i]]) son[u] = v[i];
		}
}

void DFS2(int u) {
	dfn[u] = ++ clk;
	if (son[u]) top[son[u]] = top[u], DFS2(son[u]);
	for (register int i = beg[u]; i; i = nex[i])
		if (v[i] != fa[u] && v[i] != son[u]) top[v[i]] = v[i], DFS2(v[i]);
}

struct SegmentTree {
	int tr[maxn << 2], tag[maxn << 2];
#define lc (t << 1)
#define rc (lc | 1)
#define mid ((l + r) >> 1)
	inline void pushdown(int t, int l, int r) {
		if (tag[t]) tag[lc] = ad(tag[lc], tag[t]), tag[rc] = ad(tag[rc], tag[t]), tr[lc] = ad(tr[lc], (LL)(mid - l + 1) * tag[t] % MOD), tr[rc] = ad(tr[rc], (LL)(r - mid) * tag[t] % MOD), tag[t] = 0;
	}
	void update(int t, int l, int r, int x, int y) {
		if (x <= l && r <= y) { tr[t] = ad(tr[t], (r - l + 1) % MOD), (++ tag[t]) %= MOD; return; }
		pushdown(t, l, r);
		if (x <= mid) update(lc, l, mid, x, y);
		if (y >  mid) update(rc, mid + 1, r, x, y);
		tr[t] = ad(tr[lc], tr[rc]);
	}
	int query(int t, int l, int r, int x, int y) {
		if (x <= l && r <= y) return tr[t];
		pushdown(t, l, r);
		if (y <= mid) return query(lc, l, mid, x, y);
		if (x >  mid) return query(rc, mid + 1, r, x, y);
		return ad(query(lc, l, mid, x, y), query(rc, mid + 1, r, x, y));
	}
}ST;

inline void update(int u) {
	while (top[u] != 1) ST.update(1, 1, n, dfn[top[u]], dfn[u]), u = fa[top[u]];
	ST.update(1, 1, n, 1, dfn[u]);
}

inline int query(int u) {
	int ans = 0;
	while (top[u] != 1) ans = ad(ST.query(1, 1, n, dfn[top[u]], dfn[u]), ans), u = fa[top[u]];
	ans = ad(ans, ST.query(1, 1, n, 1, dfn[u]));
	return ans;
}

int main()
{
#ifdef hany01
	freopen("bzoj3626.in", "r", stdin);
	freopen("bzoj3626.out", "w", stdout);
#endif

	static int m, l, r, z, qs, cur = 1;
	static Qry Q[maxn << 1];

	n = read(), m = read();
	For(i, 2, n) add(fa[i] = read() + 1, i);
	DFS1(1), DFS2(top[1] = 1);

	For(i, 1, m) {
		l = read() + 1, r = read() + 1, z = read() + 1;
		Q[++ qs] = (Qry){r, z, i, 1};
		if (l > 1) Q[++ qs] = (Qry){l - 1, z, i, 0};
	}
	sort(Q + 1, Q + 1 + qs);

	For(i, 1, n) for (update(i); Q[cur].i == i; ++ cur)
		ans[Q[cur].id] = ad(ans[Q[cur].id], Q[cur].ty ? query(Q[cur].u) : MOD - query(Q[cur].u));
	For(i, 1, m) printf("%d\n", ans[i]);

	return 0;
}