[HNOI2015]开店 简要题解

主席树。
推下式子，发现点的深度和好算，lca深度和不好算。
lca深度之和有个套路：先给a到根路径+1，再算b到根的和。
如果可以离线，即LNOI的LCA。本题强制在线，可持久化。
由于区间修改，使用标记永久化。
注意修改要复制全，关于和的修改有些细节：

int xiugai(int i, int j, int l, int r, int L, int R) {
	if (R <= l || r <= L) return i;
	int rt = ++sl;
	cl[rt] = cl[i];	cr[rt] = cr[i];
	he[rt] = he[i];	ld[rt] = ld[i];
	if (L <= l && r <= R) {
		he[rt] += su[j];
		ld[rt] = ld[i] + 1;
		return rt;
	}
	int m = (l + r) >> 1;
	he[rt] -= (he[cl[rt]] + he[cr[rt]]);
	cl[rt] = xiugai(cl[rt], j << 1, l, m, L, R);
	cr[rt] = xiugai(cr[rt], (j << 1) | 1, m, r, L, R);
	he[rt] += (he[cl[rt]] + he[cr[rt]]);
	return rt;
}

正常空间需要4倍mlogn：修改区间拆分2倍，上传2倍。应该跑不满。
然而本题是树剖，跑不满，所以开到\(2.5*10^7\)就行。
代码：

#include <stdio.h> 
#include <stdlib.h> 
#define M 20000010
#define ll long long
#define re register 
inline int max(int a, int b) {
	return a > b ? a: b;
}
inline int min(int a, int b) {
	return a < b ? a: b;
}
inline int read() {
	re char ch;
	while ((ch = getchar()) == '\n' || ch == ' ' || ch == '\r');
	re int jg = ch - '0';
	while ((ch = getchar()) >= '0' && ch <= '9') jg = (jg << 3) + (jg << 1) + ch - '0';
	return jg;
}
int fr[150010],ne[300010],v[300010],w[300010],bs = 0;
void addb(int a, int b, int c) {
	v[bs] = b;
	w[bs] = c;
	ne[bs] = fr[a];
	fr[a] = bs++;
}
int fa[150010],son[150010],sd[150010],cd[150010],jl[150010];
int dfs1(int u, int f) {
	fa[u] = f;
	sd[u] = sd[f] + 1;
	jl[u] = jl[f] + cd[u];
	int ma = -1,he = 1;
	son[u] = -1;
	for (int i = fr[u]; i != -1; i = ne[i]) {
		if (v[i] == f) continue;
		cd[v[i]] = w[i];
		int rt = dfs1(v[i], u);
		he += rt;
		if (rt > ma) {
			ma = rt;
			son[u] = v[i];
		}
	}
	return he;
}
int top[150010],bh[150010],tm = 0;
void dfs2(int u, int f, int tp) {
	top[u] = tp;
	bh[u] = ++tm;
	if (son[u] == -1) return;
	dfs2(son[u], u, tp);
	for (int i = fr[u]; i != -1; i = ne[i]) {
		if (v[i] != f && v[i] != son[u]) dfs2(v[i], u, v[i]);
	}
}
int js[150010],cl[M],cr[M],ld[M],sl = 0,su[600010];
ll he[M],jh[150010];
void jianshu(int i, int l, int r) {
	if (l + 1 == r) {
		su[i] = js[l];
		return;
	}
	int m = (l + r) >> 1;
	jianshu(i << 1, l, m);
	jianshu((i << 1) | 1, m, r);
	su[i] = su[i << 1] + su[(i << 1) | 1];
}
int jianshu(int l, int r) {
	int rt = ++sl;
	he[rt] = ld[rt] = 0;
	if (l + 1 == r) return rt;
	int m = (l + r) >> 1;
	cl[rt] = jianshu(l, m);
	cr[rt] = jianshu(m, r);
	return rt;
}
int build(int n) {
	for (int i = 1; i <= n; i++) js[bh[i]] = cd[i];
	jianshu(1, 1, n + 1);
	return jianshu(1, n + 1);
}
int xiugai(int i, int j, int l, int r, int L, int R) {
	if (R <= l || r <= L) return i;
	int rt = ++sl;
	cl[rt] = cl[i];	cr[rt] = cr[i];
	he[rt] = he[i];	ld[rt] = ld[i];
	if (L <= l && r <= R) {
		he[rt] += su[j];
		ld[rt] = ld[i] + 1;
		return rt;
	}
	int m = (l + r) >> 1;
	he[rt] -= (he[cl[rt]] + he[cr[rt]]);
	cl[rt] = xiugai(cl[rt], j << 1, l, m, L, R);
	cr[rt] = xiugai(cr[rt], (j << 1) | 1, m, r, L, R);
	he[rt] += (he[cl[rt]] + he[cr[rt]]);
	return rt;
}
ll getsum(int i, int j, int l, int r, int L, int R, int lh) {
	if (R <= l || r <= L) return 0;
	if (L <= l && r <= R) return 1ll * lh * su[j] + he[i];
	int m = (l + r) >> 1;
	return getsum(cl[i], j << 1, l, m, L, R, lh + ld[i]) + getsum(cr[i], (j << 1) | 1, m, r, L, R, lh + ld[i]);
}
int xiugai(int od, int x, int n) {
	while (x != 0) {
		od = xiugai(od, 1, 1, n + 1, bh[top[x]], bh[x] + 1);
		x = fa[top[x]];
	}
	return od;
}
ll getsum(int ro, int x, int n) {
	ll jg = 0;
	while (x != 0) {
		jg += getsum(ro, 1, 1, n + 1, bh[top[x]], bh[x] + 1, 0);
		x = fa[top[x]];
	}
	return jg;
}
struct SPx {
	int z,	i;
};
SPx px[150010];
int cmp(const void * a, const void * b) {
	return ((SPx * ) a) ->z - ((SPx * ) b) ->z;
}
int find(int n, int x) {
	int l = 0,	r = n;
	while (l < r) {
		int m = (l + r + 1) >> 1;
		if (px[m].z <= x) l = m;
		else r = m - 1;
	}
	return l;
}
int gen[150010];
ll getans(int L, int R, int u, int n) {
	int r = find(n, R);
	int l = find(n, L - 1);
	ll lc = getsum(gen[r], u, n) - getsum(gen[l], u, n);
	ll jg = 1ll * (r - l) * jl[u] + (jh[r] - jh[l]) - lc * 2;
	return jg;
}
void insert(int i, int n) {
	gen[i] = xiugai(gen[i - 1], px[i].i, n);
	jh[i] = jh[i - 1] + jl[px[i].i];
}
int main() {
	int n,	q,	m;
	ll la = 0;
	scanf("%d%d%d", &n, &q, &m);
	for (int i = 1; i <= n; i++) {
		fr[i] = -1;
		px[i].i = i;
		px[i].z = read();
	}
	qsort(px + 1, n, sizeof(SPx), cmp);
	for (int i = 0; i < n - 1; i++) {
		int a,		b,		c;
		a = read();
		b = read();
		c = read();
		addb(a, b, c);
		addb(b, a, c);
	}
	dfs1(1, 0);
	dfs2(1, 0, 1);
	gen[0] = build(n);
	for (int i = 1; i <= n; i++) insert(i, n);
	for (int i = 0; i < q; i++) {
		int u,		a,		b;
		u = read();
		a = read();
		b = read();
		int L = min((a + la) % m, (b + la) % m);
		int R = max((a + la) % m, (b + la) % m);
		la = getans(L, R, u, n);
		printf("%lld\n", la);
	}
	return 0;
}