peaks

给定一个无向图，点有点权边有边权

Q次询问，每次询问从点v开始只经过边权<=x的边能到达所有点中点权第k大值，无解输出-1

N<=1e5,M,Q<=5e5

建立大根kruskal重构树，每次在重构树上倍增跳父亲，跳到最浅的边权<=x的父亲

将kruskal所有叶子节点按照dfn的顺序建立一个序列，然后每次查询相当于序列区间第k大，显然可以用主席树维护

代码里为了方便只给叶子节点赋了dfn，kruskal重构树每个节点记录dfn表示他所有儿子中dfn的最小的那个，记录sz表示叶子数量

然后代码写得乱七八糟凑合着看吧

注意kruskal重构树倍增范围是所有的节点(2*n-1)不是n！！！

#include <bits/stdc++.h>
using namespace std;

struct edge
{
	int u, v, w;
} a[500010];

int n, m, q;
int h[200010], dfn[200010], dfntot;
int fa[200010][19], ds[200010], lch[200010], rch[200010], sz[200010], tmp[200010];
int val[100010];
int disc[100010];

int sb;
int root[100010];

int tree[8000010], l[8000010], r[8000010], tot;

int build_tree(int cl, int cr)
{
	tree[++tot] = 0;
	if (cl < cr)
	{
		int mid = (cl + cr) / 2;
		l[tot] = build_tree(cl, mid);
		r[tot] = build_tree(mid + 1, cr);
	}
	return tot;
}

int newtree(int oldtree, int cl, int cr, int key)
{
	int p = ++tot;
	l[p] = l[oldtree];
	r[p] = r[oldtree];
	tree[p] = tree[oldtree] + 1;
	if (cl < cr)
	{
		if (key > (cl + cr) / 2)
			r[p] = newtree(r[oldtree], (cl + cr) / 2 + 1, cr, key);
		else
			l[p] = newtree(l[oldtree], cl, (cl + cr) / 2, key);
	}
	return p;
}

int query(int tree2, int tree1, int cl, int cr, int key)
{
	if (cl >= cr)
		return cl;
	int sz = tree[l[tree2]] - tree[l[tree1]];
	if (sz >= key)
		return query(l[tree2], l[tree1], cl, (cl + cr) / 2, key);
	else
		return query(r[tree2], r[tree1], (cl + cr) / 2 + 1, cr, key - sz);
}

void jianshu()
{
	root[0] = build_tree(1, sb);
	for (int i = 1; i <= n; i++)
		root[i] = newtree(root[i - 1], 1, sb, val[i]);
}

int kth(int l, int r, int k)
{
	return query(root[r], root[l - 1], 1, sb, r - l + 2 - k);
	return 233;
}

int getf(int x)
{
	return ds[x] == x ? x : ds[x] = getf(ds[x]);
}

void dfs(int x)
{
	if (x <= n)
	{
		dfn[x] = ++dfntot;
		sz[x] = 1;
		val[dfn[x]] = h[x];
		return;
	}
	dfs(lch[x]);
	dfs(rch[x]);
	dfn[x] = dfn[lch[x]];
	sz[x] = sz[lch[x]] + sz[rch[x]];
}

int main()
{
	scanf("%d%d%d", &n, &m, &q);
	for (int i = 1; i <= n; i++)
		scanf("%d", &h[i]);
	for (int i = 1; i <= m; i++)
		scanf("%d%d%d", &a[i].u, &a[i].v, &a[i].w);
	sort(a + 1, a + 1 + m, [](const edge &a, const edge &b) {return a.w < b.w;});
	for (int i = 1; i <= n; i++)
		ds[i] = i;
	int tot = n;
	for (int i = 1; i <= m; i++)
	{
		if (getf(a[i].u) != getf(a[i].v))
		{
			int p = ++tot, p1 = getf(a[i].u), p2 = getf(a[i].v);
			tmp[p] = a[i].w;
			ds[p1] = ds[p2] = ds[p] = p;
			fa[p1][0] = fa[p2][0] = p;
			lch[p] = p1;
			rch[p] = p2;
		}
	}
	dfs(tot);
	for (int j = 1; j <= 18; j++)
		for (int i = 1; i <= tot; i++)
			fa[i][j] = fa[fa[i][j - 1]][j - 1];
	// for (int i = 1; i <= n; i++)
		// printf("%d%c", val[i], i == n ? '\n' : ' ');
	tmp[0] = 0x3f3f3f3f;
	for (int i = 1; i <= n; i++)
		disc[i] = val[i];
	sort(disc + 1, disc + 1 + n);
	sb = unique(disc + 1, disc + 1 + n) - disc - 1;
	for (int i = 1; i <= n; i++)
		val[i] = lower_bound(disc + 1, disc + 1 + sb, val[i]) - disc;
	// for (int i = 1; i <= n; i++)
		// printf("%d%c", val[i], i == n ? '\n' : ' ');
	jianshu();
	for(int i = 1; i <= q; i++)
	{
		int v, x, k;
		scanf("%d%d%d", &v, &x, &k);
		for (int j = 18; j >= 0; j--)
			if (tmp[fa[v][j]] <= x)
				v = fa[v][j];
		if (sz[v] < k)
			puts("-1");
		else
			printf("%d\n", disc[kth(dfn[v], dfn[v] + sz[v] - 1, k)]);
	}
	return 0;
}