这题有点像【雨天的尾巴】【永无乡】的结合版本，树上差分，线段树合并，权值线段树查找第$k$大。

对于操作$i$，我们可以对$u->v$路径上的点，$i$的权值加上$1$，然后线段树合并，查找第$k$大就好了。

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10;int head[N], to[N << 1], nex[N << 1], cnt = 1;int n, m, k, ans[N], root[N], value[N];int fa[N], top[N], dep[N], son[N], sz[N];int ls[N * 60], rs[N * 60], sum[N * 60], tot;void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}void dfs1(int rt, int f) {fa[rt] = f, dep[rt] = dep[f] + 1, sz[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];}}
}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;
}void push_up(int rt) {sum[rt] = sum[ls[rt]] + sum[rs[rt]];
}void update(int &rt, int l, int r, int x, int value) {if (!rt) {rt = ++tot;}if (l == r) {sum[rt] += value;return ;}int mid = (l + r) >> 1;if (x <= mid) {update(ls[rt], l, mid, x, value);}if (x > mid) {update(rs[rt], mid + 1, r, x, value);}push_up(rt);
}int merge(int x, int y, int l, int r) {if (x == 0 || y == 0) {return x | y;}if (l == r) {sum[x] += sum[y];return x;}int mid = (l + r) >> 1;ls[x] = merge(ls[x], ls[y], l, mid);rs[x] = merge(rs[x], rs[y], mid + 1, r);push_up(x);return x;
}int find_k_th(int rt, int l, int r, int k) {if (l == r) {return l;}int mid = (l + r) >> 1;if (k > sum[ls[rt]]) {return find_k_th(rs[rt], mid + 1, r, k - sum[ls[rt]]);}return find_k_th(ls[rt], l, mid, k);
}void dfs(int rt, int fa) {for (int i = head[rt]; i; i = nex[i]) {if (to[i] == fa) {continue;}dfs(to[i], rt);root[rt] = merge(root[rt], root[to[i]], 1, n);}if (k <= sum[root[rt]]) {ans[rt] = find_k_th(root[rt], 1, n, sum[root[rt]] - k + 1);}
}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, &k);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 u, v, c;scanf("%d %d %d", &u, &v, &value[i]);int f = lca(u, v), ff = fa[f];update(root[u], 1, n, i, 1);update(root[v], 1, n, i, 1);update(root[f], 1, n, i, -1);if (ff) {update(root[ff], 1, n, i, -1);}}dfs(1, 0);for (int i = 1; i <= n; i++) {printf("%d%c", value[ans[i]], i == n ? '\n' : ' ');}return 0;
}

异或操作，容易想到拆位，然后依次算贡献，然后注意同一颗子树上相同的值如果出现了偶数次要相消，线段树合并一下就好了。

#include <bits/stdc++.h>using namespace std;typedef pair<int, int> pii;
typedef long long ll;const int N = 1e5 + 10;int head[N], to[N << 1], nex[N << 1], cnt = 1;int n, m, value[N], root[N];int ls[N * 33], rs[N * 33], sum[N * 33], num[N * 33][17], tot;ll ans[N];vector<pii> ask[N];void push_up(int rt) {sum[rt] = sum[ls[rt]] + sum[rs[rt]];for (int i = 0; i <= 16; i++) {num[rt][i] = num[ls[rt]][i] + num[rs[rt]][i];}
}void update(int &rt, int l, int r, int x) {if (!rt) {rt = ++tot;}if (l == r) {if (sum[rt]) {sum[rt] = 0;for (int i = 0; i <= 16; i++) {num[rt][i] = 0;}}else {sum[rt] = 1;for (int i = 0; i <= 16; i++) {num[rt][i] = x >> i & 1;}}return ;}int mid = l + r >> 1;if (x <= mid) {update(ls[rt], l, mid, x);}else {update(rs[rt], mid + 1, r, x);}push_up(rt);return ;
}int merge(int x, int y, int l, int r) {if (x == 0 || y == 0) {return x | y;}if (l == r) {for (int i = 0; i <= 16; i++) {num[x][i] ^= num[y][i];}sum[x] ^= sum[y];return x;}int mid = l + r >> 1;ls[x] = merge(ls[x], ls[y], l, mid);rs[x] = merge(rs[x], rs[y], mid + 1, r);push_up(x);return x;
}void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}ll query(int rt, int l, int r, int L, int R, int value) {if (!rt) {return 0;}if (l >= L && r <= R) {ll ans = 0;for (int i = 0; i <= 16; i++) {if (value >> i & 1) {ans += (1ll << i) * (sum[rt] - num[rt][i]);}else {ans += (1ll << i) * num[rt][i];}}return ans;}int mid = l + r >> 1;ll ans = 0;if (L <= mid) {ans += query(ls[rt], l, mid, L, R, value);}if (R > mid) {ans += query(rs[rt], mid + 1, r, L, R, value);}return ans;
}void dfs(int rt, int fa) {for (int i = head[rt]; i; i = nex[i]) {if (to[i] == fa) {continue;}dfs(to[i], rt);root[rt] = merge(root[rt], root[to[i]], 1, n);}update(root[rt], 1, n, value[rt]);for (auto it : ask[rt]) {if (it.second != n) {ans[it.first] = query(root[rt], 1, n, it.second + 1, n, it.second);}}
}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", &value[i]);}for (int i = 1; i < n; i++) {int x, y;scanf("%d %d", &x, &y);add(x, y);add(y, x);}for (int i = 1; i <= m; i++) {int u, x;scanf("%d %d", &u, &x);ask[u].push_back(make_pair(i, x));}dfs(1, 0);for (int i = 1; i <= m; i++) {printf("%lld\n", ans[i]);}return 0;
}