给定一棵有$n$个节点的树，点有点权，有两种操作：① 修改某个点的点权，② 查询两点路径间的点权第$k$大。

给定$u,v$，选定$1$号节点为根节点，设$inf(x)$表示从根节点到点$x$的信息，两点间的信息可以描述为$inf(u)+inf(v)-inf(lca(u,v))-fa(inf(lca(u,v)))$，

由此我们可以把树上每个点拆成$[l,r,x]$，$l$为这个点进入$dfs$的$dfs$序，$r$为这个点走出$dfs$的$dfs$序，

我们要查找$inf(u)$上的信息就只要查找有多少个$[l,r,x]$满足$l\le dfn(u)\le r$，树状数组区间更新，单点查询就好了。

对于每个询问可以写成$u,v,lca(u,v),fa(lca(u,v))$，我们把树上的点，修改操作，询问操作放到数组中整体二分一下即可。

（感觉代码好像比树套树好写一点点……）

#include <bits/stdc++.h>using namespace std;const int N = 2e5 + 10;int head[N], to[N], nex[N], cnt = 1;int a[N], ans[N], n, m;int son[N], fa[N], l[N], dep[N], r[N], sz[N], top[N], sum[N], tot;struct Res {int l, r, f, ff, x, id, type;
}q[N], q1[N], q2[N];void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void dfs1(int rt, int f) {dep[rt] = dep[f] + 1, fa[rt] = f, sz[rt] = 1, l[rt] = ++tot;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[to[i]] > sz[son[rt]]) {son[rt] = to[i];}}r[rt] = tot;
}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;
}inline int lowbit(int x) {return x & (-x);
}void update(int rt, int v) {while (rt <= N) {sum[rt] += v;rt += lowbit(rt);}
}int query(int rt) {int ans = 0;while (rt) {ans += sum[rt];rt -= lowbit(rt);}return ans;
}void solve(int l, int r, int L, int R) {if (l > r || L > R) {return ;}if (l == r) {for (int i = L; i <= R; i++) {if (q[i].type == 2) {ans[q[i].id] = l;}}return ;}int mid = l + r >> 1, cnt1 = 0, cnt2 = 0;for (int i = L; i <= R; i++) {if (q[i].type == 1) {if (q[i].x <= mid) {int l = q[i].l, r = q[i].r;update(l, q[i].id), update(r + 1, -q[i].id);q1[++cnt1] = q[i];}else {q2[++cnt2] = q[i];}}else {int cur = query(q[i].l) + query(q[i].r) - query(q[i].f) - query(q[i].ff);if (cur >= q[i].x) {q1[++cnt1] = q[i];}else {q[i].x -= cur;q2[++cnt2] = q[i];}}}for (int i = 1; i <= cnt1; i++) {if (q1[i].type == 1) {int l = q1[i].l, r = q1[i].r;update(l, -q1[i].id), update(r + 1, q1[i].id);}}for (int i = 1; i <= cnt1; i++) {q[L + i - 1] = q1[i];}for (int i = 1; i <= cnt2; i++) {q[L + cnt1 + i - 1] = q2[i];}solve(l, mid, L, L + cnt1 - 1), solve(mid + 1, r, L + cnt1, R);
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);scanf("%d %d", &n, &m);for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);}for (int i = 1, x, y; i < n; i++) {scanf("%d %d", &x, &y);add(x, y);add(y, x);}dfs1(1, 0);dfs2(1, 1);for (int i = 1; i <= m; i++) {ans[i] = 1000000000;}int num = 0;for (int i = 1; i <= n; i++) {q[++num] = {l[i], r[i], 0, 0, a[i], 1, 1};}for (int i = 1, k, x, y; i <= m; i++) {scanf("%d %d %d", &k, &x, &y);if (!k) {q[++num] = {l[x], r[x], 0, 0, a[x], -1, 1};a[x] = y;q[++num] = {l[x], r[x], 0, 0, a[x], 1, 1};}else {int f = lca(x, y), ff = fa[f], sum = dep[x] + dep[y] - dep[f] - dep[ff];if (sum >= k) {q[++num] = {l[x], l[y], l[f], l[ff], sum - k + 1, i, 2};}else {ans[i] = -1;}}}solve(0, 100000000, 1, num);for (int i = 1; i <= m; i++) {if (ans[i] != 1000000000) {if (ans[i] == -1) {puts("invalid request!");}else {printf("%d\n", ans[i]);}}}return 0;
}