并查集+map

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) (vv).begin(), (vv).end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e5 + 7;
int fa[N], cnt[N], ans[N];
map<int, int> mp[N];

int find(int x) {
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}

int main() {
    js;
    int n, m, q;
    while (cin >> n >> m >> q) {
        for (int i = 1; i <= n; ++i)    mp[i].clear(), fa[i] = i;
        fill(cnt + 1, cnt + m + 1, 0);
        fill(ans + 1, ans + 1 + m, -1); //初始化-1
        for (int i = 1; i <= n; ++i) {
            int x;    cin >> x;
            ++cnt[x];
            ++mp[i][x];
        }
        for (int i = 1; i <= q; ++i) {
            int u, v;    cin >> u >> v;
            u = find(u), v = find(v);
            if (mp[u].size() < mp[v].size())    swap(u, v);
            fa[v] = u;  //小的合并到大的里面去
            for (auto it : mp[v]) {
                mp[u][it.first] += it.second;
                if (mp[u][it.first] == cnt[it.first] and ans[it.first] == -1) //都已连线并且当前ans是没有求解的状态
                    ans[it.first] = i;
            }
            mp[v].clear();
        }
        for (int i = 1; i <= m; ++i)
            if (cnt[i] == 1)    cout << 0 << endl;
            else if (cnt[i] == -1)    cout << -1 << endl;
            else  cout << ans[i] << endl;
    }
    return 0;
}