wa 这题重写了一遍过了 我第一次写的什么狗屎啊

题目 ： https://nanti.jisuanke.com/t/38229

如下图

对每个 树节点 以他向后面每个链 建主席树
这样只要跑lca 就能区间快查k大了

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 100000 + 10;

int head[maxn], depth[maxn], cnt;
int shu;
int dis[maxn << 1];
int nxt[maxn << 1], to[maxn << 1];

int fa[maxn][30], lg[maxn];
int n, m, k, q;

void ade(int a, int b, int c) {
	to[++cnt] = b;
	nxt[cnt] = head[a];
	dis[cnt] = c;
	head[a] = cnt;
}

struct node {
	int lc, rc;
	int val;
} tree[maxn * 20];

int tot, root[maxn];
int b[maxn];

int build(int l, int r) {
	int p = ++ tot;
	if(l == r) {
		tree[p].val = 0;
		return p;
	}
	int mid = l + r >> 1;
	tree[p].lc = build(l, mid);
	tree[p].rc = build(mid + 1, r);
	tree[p].val = tree[tree[p].lc].val + tree[tree[p].rc].val;
	return p;
}

int ins(int now, int l, int r, int pos, int val) {
	int p = ++ tot;
	tree[p] = tree[now];
	if(l == r) {
		tree[p].val += val ;
		return p;
	}
	int mid = l + r >> 1;
	if(pos <= mid)
		tree[p].lc = ins(tree[now].lc, l, mid, pos, val);
	else
		tree[p].rc = ins(tree[now].rc, mid + 1, r, pos, val);
	tree[p].val = tree[tree[p].lc].val + tree[tree[p].rc].val;
	return p;
}

int ask(int p, int q, int l, int r, int L, int R) {
	if(R < L) return 0;
	if(L <= l && r <= R)
		return tree[p].val - tree[q].val;

	int mid = l + r >> 1;
	int res = 0;
	if(L <= mid)
		res += ask(tree[p].lc, tree[q].lc, l, mid, L, R);
	if(R > mid)
		res += ask(tree[p].rc, tree[q].rc, mid + 1, r, L, R);
	return res;
}

int getid(int x) {
	int res = upper_bound(b + 1, b + shu, x) - b - 1;
	return res;
}

void dfs(int x, int pre) {
	depth[x] = depth[pre] + 1;
	fa[x][0] = pre;
	for(int i = 1; (1 << i) <= depth[x]; i++)
		fa[x][i] = fa[fa[x][i - 1]][i - 1];
	for(int i = head[x]; i; i = nxt[i])
		if(to[i] != pre) {
			root[to[i]] = ins(root[x], 1, shu, getid(dis[i]), 1);
			dfs(to[i], x);
		}
}

int lca(int x, int y) {
	if(depth[x] < depth[y])
		swap(x, y);
	while(depth[x] > depth[y])
		x = fa[x][lg[depth[x] - depth[y]] - 1];
	if(x == y)
		return x;
	for(int k = lg[depth[x]] - 1; k >= 0; k--) {
		if(fa[x][k] != fa[y][k]) {
			x = fa[x][k], y = fa[y][k];
		}
	}
	return fa[x][0];
}

void lsh() {
	sort(b + 1, b + n);
	shu = unique(b + 1, b + n) - b;
}

int main() {
	for(int i = 1; i < maxn; i++)
		lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);

	scanf("%d %d", &n, &m);
	for(int i = 1; i < n; i ++ ) {
		int st, ed, dis;
		cin >> st >> ed >> dis;
		ade(st, ed, dis),ade(ed, st, dis);
		b[i] = dis;
	}

	lsh();

	root[1] = build(1, shu);

	dfs(1,0);

	for(int i = 1; i <= m; i ++) {
		int x, y, k;
		scanf( "%d %d %d" , &x, &y, &k);
		k = getid(k);
		int LCA = lca(x, y);
		printf( "%d\n",\
		        ask(root[y], root[LCA], 1, shu, 1, k)\
		        + ask(root[x], root[LCA], 1, shu, 1, k)\
		      );
	}
	return 0;
}