1805E - There Should Be a Lot of Maximums - CodeForces Solution

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2000'010;
const ll mod = 1000'000'000LL;
vector<int> adj[N];
int mx = 0, a[N], in[N], out[N], rs[N], cs = 0;
map<int, int> cnt[2];
map<pair<int,int>,int> id;
set<int> s[2];
void add(int x, int p)
{
    cnt[p][a[x]]++;
    if(cnt[p][a[x]] == 2)
        s[p].insert(-a[x]);
}
void rmv(int x, int p)
{
    if(cnt[p][a[x]] == 2)
        s[p].erase(-a[x]);
    cnt[p][a[x]]--;
}
int getans(int x)
{
    if(s[x].empty())
        return 0;
    return -*s[x].begin();
}
void dfs(int u, int p, int bad)
{
    in[u]=cs++;
    if(u == bad)
        bad = -1;
    if(bad != -1){
        add(u, 0);
    }
    else
        add(u, 1);
    for(int v: adj[u])
    {
        if(v == p)
            continue;
        dfs(v, u, bad);
    }
    out[u]=cs++;
}
void swp(int u, int p)
{
    // cout << u << '\n';
    rmv(u, 0);
    add(u, 1);
    for(int v: adj[u])
    {
        if(v == p)
            continue;
        swp(v, u);
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin >> n;
    for(int i = 0; i < n-1; i++)
    {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
        id[{u,v}] = id[{v,u}] = i;
        rs[i] = -1;
    }
    for(int i = 1; i <= n; i++)
    {
        cin >> a[i];
        cnt[0][a[i]]++;
    }
    for(int i = 1; i <= n; i++)
    {
        if(cnt[0][a[i]] > 1)
            mx = max(mx, a[i]);
    }
    if(cnt[0][mx] > 2 || mx == 0)
    {
        for(int i = 0; i < n-1; i++)
            cout << mx << '\n';
        return 0;
    }
    cnt[0].clear();
    int u, v;
    u = -1;
    for(int i = 1; i <= n; i++)
    {
        if(a[i] == mx)
        {
            if(u == -1)
                u = i;
            else
                v = i; 
        }
    }
    // cout << u << ' ' << v << '\n';
    dfs(u, u, v);
    int x = v, px = v;
    while(x != u)
    {
        for(int at: adj[x])
        {
            if(in[at] <= in[v] && out[v] <= out[at] && px != at)
            {
                px = x;
                x = at;
                break;
            }
        }
        rs[id[{px,x}]] = max(getans(1),getans(0));
        // cout << x << ' ' << px  << " " << rs[id[{px,x}]]<< " << \n";
        for(int at: adj[x])
        {
            if(!(in[at]<= in[v] && out[v] <= out[at]))
            {
                swp(at, x);
                // break;
            }
        }
        rmv(x,0);
        add(x,1);
    }
    for(int i = 0; i < n-1; i++)
    {
        if(rs[i] == -1)
            cout << mx << '\n';
        else
            cout << rs[i] << '\n';
    }
    return 0;
}