我们要求从一个点出发经过困难值小于等于$x$的路径所能到达的山峰中第$k$高的是什么。

考虑按照边权升序，建议$kruskal$重构树，然后倍增向上跳，找到困难值小于等于$x$的深度最小的节点$u$，

那么我们只要在$u$的子树中询问第$k$大即可，所以可以用主席树来写，依照$dfs$序，对每个节点建立一颗主席树，然后在主席树上查找第$k$大即可。

#include <bits/stdc++.h>using namespace std;const int N = 5e5 + 10;int head[N], to[N], nex[N], cnt = 1;int n, m, q, nn, a[N], ff[N], value[N], h[N];int fa[N][21], l[N], r[N], rk[N], tot;int root[N], ls[N << 7], rs[N << 7], sum[N << 7], num;struct Res {int u, v, w;bool operator < (const Res &t) const {return w < t.w;}
}edge[N];void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void update(int &rt, int pre, int l, int r, int x, int v) {rt = ++num;ls[rt] = ls[pre], rs[rt] = rs[pre], sum[rt] = sum[pre] + v;if (l == r) {return ;}int mid = l + r >> 1;if (x <= mid) {update(ls[rt], ls[pre], l, mid, x, v);}else {update(rs[rt], rs[pre], mid + 1, r, x, v);}
}int query(int L, int R, int l, int r, int k) {if (l == r) {return l;}int res = sum[ls[R]] - sum[ls[L]], mid = l + r >> 1;if (res >= k) {return query(ls[L], ls[R], l, mid, k);}else {return query(rs[L], rs[R], mid + 1, r, k - res);}
}int find(int rt) {return rt == ff[rt] ? rt : ff[rt] = find(ff[rt]);
}void dfs(int rt, int f) {fa[rt][0] = f, l[rt] = ++tot, rk[tot] = rt;for (int i = 1; i <= 20; i++) {fa[rt][i] = fa[fa[rt][i - 1]][i - 1];}for (int i = head[rt]; i; i = nex[i]) {if (to[i] == f) {continue;}dfs(to[i], rt);}r[rt] = tot;
}void kruskal() {for (int i = 1; i < N; i++) {ff[i] = i;}sort(edge + 1, edge + 1 + m);for (int i = 1, cur = 1; i <= m && cur < n; i++) {int u = find(edge[i].u), v = find(edge[i].v);if (u != v) {cur++, nn++;ff[u] = nn, ff[v] = nn;value[nn] = edge[i].w;add(nn, u), add(nn, v);if (u <= n) {value[u] = edge[i].w;}if (v <= n) {value[v] = edge[i].w;}}}dfs(nn, 0);
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d %d %d", &n, &m, &q);for (int i = 1; i <= n; i++) {scanf("%d", &h[i]);a[i] = h[i];}nn = n;for (int i = 1; i <= m; i++) {scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].w);}kruskal();int maxn = n;sort(a + 1, a + 1 + maxn);maxn = unique(a + 1, a + 1 + maxn) - (a + 1);for (int i = 1; i <= n; i++) {h[i] = lower_bound(a + 1, a + 1 + maxn, h[i]) - a; }for (int i = 1; i <= nn; i++) {root[i] = root[i - 1];if (rk[i] <= n) {update(root[i], root[i], 1, maxn, h[rk[i]], 1);}}for (int i = 1, u, x, k, last_ans = 0, res; i <= q; i++) {scanf("%d %d %d", &u, &x, &k);if (last_ans != -1) {u ^= last_ans, x ^= last_ans, k ^= last_ans;}for (int j = 20; j >= 0; j--) {if (fa[u][j] && value[fa[u][j]] <= x) {u = fa[u][j];}}res = sum[root[r[u]]] - sum[root[l[u] - 1]];last_ans = res < k ? -1 : a[query(root[l[u] - 1], root[r[u]], 1, maxn, res - k + 1)];printf("%d\n", last_ans);}return 0;
}