제출 #931678

#제출 시각아이디문제언어결과실행 시간메모리
931678LOLOLORegions (IOI09_regions)C++17
100 / 100
3360 ms72708 KiB
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

#define           f     first
#define           s     second
#define           pb    push_back
#define           ep    emplace
#define           eb    emplace_back
#define           lb    lower_bound
#define           ub    upper_bound
#define       all(x)    x.begin(), x.end()
#define      rall(x)    x.rbegin(), x.rend()
#define   uniquev(v)    sort(all(v)), (v).resize(unique(all(v)) - (v).begin())
#define     mem(f,x)    memset(f , x , sizeof(f))
#define        sz(x)    (int)(x).size()
#define  __lcm(a, b)    (1ll * ((a) / __gcd((a), (b))) * (b))
#define          mxx    *max_element
#define          mnn    *min_element
#define    cntbit(x)    __builtin_popcountll(x)
#define       len(x)    (int)(x.length())

const int N = 2e5 + 100;
const int lim = 500;

vector <int> lst[N], ed[N], all, e[N];
map <int, int> ans[N];
int c[N], in[N], ou[N], timer = 1, cnt[N];

void dfs(int u) {
    cnt[c[u]]++;
    in[u] = ++timer;

    for (auto x : ed[u]) {
        dfs(x);
    }

    cnt[c[u]]--;
    for (auto x : all) {
        ans[x][c[u]] += cnt[x];
    }

    ou[u] = timer;
}

int main() {
    int n, r, q;
    cin >> n >> r >> q;
    cin >> c[1];

    for (int i = 2; i <= n; i++) {
        int x;
        cin >> x;
        cin >> c[i];
        ed[x].pb(i);
    }

    for (int i = 1; i <= n; i++) {
        lst[c[i]].pb(i);
    }

    for (int i = 1; i <= r; i++) {
        if (sz(lst[i]) >= lim)
            all.pb(i);
    }

    dfs(1);

    for (int i = 1; i <= n; i++) {
        e[c[i]].pb(in[i]);
    }

    for (int i = 1; i <= r; i++) {
        sort(all(e[i]));
    }

    for (int i = 1; i <= q; i++) {
        int r1, r2;
        cin >> r1 >> r2;
        if (sz(lst[r1]) >= lim) {
            cout << ans[r1][r2] << '\n';
        } else {
            int cnt = 0;
            for (auto x : lst[r1]) {
                int l = lower_bound(all(e[r2]), in[x]) - e[r2].begin(), r = upper_bound(all(e[r2]), ou[x]) - e[r2].begin() - 1;
                cnt += (r - l + 1);
            }

            cout << cnt << '\n';
        }
    }

    return 0;
}

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