题目链接：学霸大帅哥zyh

dsu on tree，用线段树维护答案，每次修改的时候只会改变一个数，所以只会变一条链，非常好写。复杂度 \(O(n\log^2 n)\)。题是好题，就是数据太水了，\(O(n^2\log n)\) 的假做法都能过。

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define endl '\n'
using ll = long long;
using pii = pair<int, int>;
struct node {
    int id, cnt;
    bool operator<(const node &other) const { return cnt > other.cnt; }
};
const int maxn = 1e5 + 5;
multiset<node> s;
vector<pii> g[maxn];
int tot[maxn], h[maxn];
int sz[maxn], now[maxn];
pii son[maxn];
int ans[maxn], vis[maxn];
namespace sgt {
#define ls p << 1
#define rs p << 1 | 1
int t[maxn << 2], id[maxn << 2];
void push_up(int p) {
    t[p] = max(t[ls], t[rs]);
    if (t[ls] != t[rs]) {
        id[p] = t[ls] > t[rs] ? id[ls] : id[rs];
    } else {
        id[p] = t[p] > 0 ? max(id[ls], id[rs]) : -1;
    }
}
void build(int p, int l, int r) {
    if (l == r) {
        id[p] = l;
        return;
    }
    int mid = (l + r) >> 1;
    build(ls, l, mid);
    build(rs, mid + 1, r);
    push_up(p);
}
void modify(int p, int x, int l, int r, int k) {
    if (l == r) {
        assert(l == x);
        t[p] = k;
        return;
    }
    int mid = (l + r) >> 1;
    if (x <= mid)
        modify(ls, x, l, mid, k);
    if (x > mid)
        modify(rs, x, mid + 1, r, k);
    push_up(p);
}
int query() { return id[1]; }
}; // namespace sgt
void dfs0(int u, int f) {
    sz[u] = 1;
    for (auto &&p : g[u]) {
        auto &v = p.first;
        if (v == f)
            continue;
        dfs0(v, u);
        sz[u] += sz[v];
        if (sz[v] > sz[son[u].first])
            son[u] = p;
    }
}
int S;
void work(int u, int f, int val) {
    now[h[u]] += val;
    int t = min(now[h[u]], tot[h[u]] - now[h[u]]);
    sgt::modify(1, h[u], 1, 100000, t);
    for (auto &&p : g[u]) {
        auto &v = p.first;
        if (v == f || v == S)
            continue;
        work(v, u, val);
    }
}
map<pii, int> mp;
void dfs(int u, int f, int op) {
    for (auto &&p : g[u]) {
        auto &v = p.first;
        if (v == f || v == son[u].first)
            continue;
        dfs(v, u, 0);
    }
    if (son[u].first) {
        dfs(son[u].first, u, 1);
        S = son[u].first;
    }
    work(u, f, 1);
    S = 0;
    ans[mp[{f, u}]] = sgt::query();
    if (!op) {
        work(u, f, -1);
    }
}
void solve() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> h[i];
        tot[h[i]]++;
    }
    for (int i = 1, u, v; i < n; ++i) {
        cin >> u >> v;
        g[u].pb({v, i}), g[v].pb({u, i});
        mp[{v, u}] = i, mp[{u, v}] = i;
    }
    dfs0(1, 0);
    sgt::build(1, 1, 100000);
    dfs(1, 0, 1);
    for (int i = 1; i < n; ++i) {
        cout << ans[i] << endl;
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    // cin >> T;
    while (T--) {
        solve();
    }
}