每次询问两点间的第$k$大，考虑建立$n$棵主席树，每棵主席树记录的是从根节点到当前节点的状态，

由于主席树是一种类似前缀和的数据结构，可以使用差分的方式得到$(u,v)$点对之间的主席树为$u+v-lca(u,v)-fa(lca(u,v))$，

即这一段路径上的信息可用这四棵主席树加减得到，但是这样只能处理静态问题，这里还增加了修改操作。

利用$dfs$序在树状数组上建立主席树，当这个点进栈的时候，把这个点的信息在树状数组上的主席树上进行$+1$，当这个点出栈的时候将这个点的信息在树状数组上的主席树上进行$-1$，所以对某个点查询的时候我们查询树状数组上的主席树，其所有值加起来也就是从根节点到当前节点的信息，然而我们把修改操作优化成了$\log n\times \log n$，整体复杂度也就是$n\log n\times \log n$。

#include <bits/stdc++.h>using namespace std;const int N = 8e4 + 10, maxn = 100000000;int a[N], n, m;int head[N], to[N << 1], nex[N << 1], cnt = 1;int son[N], sz[N], l[N], r[N], dep[N], top[N], fa[N], tot;int root[N], sum[N << 8], ls[N << 8], rs[N << 8], num;int A[N], B[N], cnt1, cnt2;inline int lowbit(int x) {return x & (-x);
}inline void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void update(int &rt, int l, int r, int x, int v) {if (!rt) {rt = ++num;}sum[rt] += v;if (l == r) {return ;}int mid = l + r >> 1;if (x <= mid) {update(ls[rt], l, mid, x, v);}else {update(rs[rt], mid + 1, r, x, v);}
}void dfs1(int rt, int f) {fa[rt] = f, l[rt] = ++tot, sz[rt] = 1, dep[rt] = dep[f] + 1;for (int i = l[rt]; i <= n; i += lowbit(i)) {update(root[i], 0, maxn, a[rt], 1);}for (int i = head[rt]; i; i = nex[i]) {if (to[i] == f) {continue;}dfs1(to[i], rt);sz[rt] += sz[to[i]];if (!son[rt] || sz[son[rt]] < sz[to[i]]) {son[rt] = to[i];}}r[rt] = tot;for (int i = r[rt] + 1; i <= n; i += lowbit(i)) {update(root[i], 0, maxn, a[rt], -1);}
}void dfs2(int rt, int tp) {top[rt] = tp;if (!son[rt]) {return ;}dfs2(son[rt], tp);for (int i = head[rt]; i; i = nex[i]) {if (to[i] == fa[rt] || to[i] == son[rt]) {continue;}dfs2(to[i], to[i]);}
}int lca(int x, int y) {while (top[x] != top[y]) {if (dep[top[x]] < dep[top[y]]) {swap(x, y);}x = fa[top[x]];}return dep[x] < dep[y] ? x : y;
}int query(int l, int r, int k) {if (l == r) {return l;}int cur = 0, mid = l + r >> 1;for (int i = 1; i <= cnt1; i++) {cur += sum[ls[A[i]]];}for (int i = 1; i <= cnt2; i++) {cur -= sum[ls[B[i]]];}if (cur >= k) {for (int i = 1; i <= cnt1; i++) {A[i] = ls[A[i]];}for (int i = 1; i <= cnt2; i++) {B[i] = ls[B[i]];}return query(l, mid, k);}else {for (int i = 1; i <= cnt1; i++) {A[i] = rs[A[i]];}for (int i = 1; i <= cnt2; i++) {B[i] = rs[B[i]];}return query(mid + 1, r, k - cur);}
}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", &n, &m);for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);}for (int i = 1; i < n; i++) {int x, y;scanf("%d %d", &x, &y);add(x, y);add(y, x);}dfs1(1, 0);dfs2(1, 1);for (int i = 1; i <= m; i++) {int k, u, v;scanf("%d %d %d", &k, &u, &v);if (!k) {for (int i = l[u]; i <= n; i += lowbit(i)) {update(root[i], 0, maxn, a[u], -1);}for (int i = r[u] + 1; i <= n; i += lowbit(i)) {update(root[i], 0, maxn, a[u], 1);}a[u] = v;for (int i = l[u]; i <= n; i += lowbit(i)) {update(root[i], 0, maxn, a[u], 1);}for (int i = r[u] + 1; i <= n; i += lowbit(i)) {update(root[i], 0, maxn, a[u], -1);}}else {int f = lca(u, v), ff = fa[f], total = dep[u] + dep[v] - dep[f] - dep[ff];if (total < k) {puts("invalid request!");}else {cnt1 = cnt2 = 0;for (int i = l[u]; i; i -= lowbit(i)) {A[++cnt1] = root[i];}for (int i = l[v]; i; i -= lowbit(i)) {A[++cnt1] = root[i];}for (int i = l[f]; i; i -= lowbit(i)) {B[++cnt2] = root[i];}for (int i = l[ff]; i; i -= lowbit(i)) {B[++cnt2] = root[i];}k = total - k + 1;printf("%d\n", query(0, maxn, k));}}}return 0;
}