Submission #518348

#	Time	Username	Problem	Language	Result	Execution time	Memory
518348		Alex_tz307	Regions (IOI09_regions)	C++17	100 / 100	4529 ms	31504 KiB

regions

#include <bits/stdc++.h>

using namespace std;

const int kN = 2e5;
const int kR = 25 * 1e3;
const int kBuck = 450;
int timer, r[kN], in[kN], out[kN], ind[kR], cnt[kBuck], sum[1 + kN], root1[kBuck][kR], root2[kBuck][kR];
vector<int> g[kN], R[kR], pos[kR], large;
bitset<kN> isLarge;

void dfs1(int u) {
  in[u] = ++timer;
  pos[r[u]].emplace_back(in[u]);
  for (int v : g[u])
    dfs1(v);
  out[u] = timer;
}

void dfs2(int u) {
  for (int region : large) {
    root1[ind[region]][r[u]] += cnt[ind[region]];
  }
  if (isLarge[r[u]]) {
    ++cnt[ind[r[u]]];
  }
  for (int v : g[u]) {
    dfs2(v);
  }
  if (isLarge[r[u]]) {
    --cnt[ind[r[u]]];
  }
}

int solve(int r1, int r2) {
  int ans = 0;
  for (int u : R[r1]) {
    int st = lower_bound(pos[r2].begin(), pos[r2].end(), in[u]) - pos[r2].begin();
    int dr = upper_bound(pos[r2].begin(), pos[r2].end(), out[u]) - pos[r2].begin();
    ans += dr - st;
  }
  return ans;
}

int main() {
  int n, m, Q;
  cin >> n >> m >> Q;
  for (int v = 0; v < n; ++v) {
    if (v == 0) {
      cin >> r[0];
      --r[0];
      R[r[0]].emplace_back(0);
      continue;
    }
    int u;
    cin >> u >> r[v];
    --r[v];
    g[u - 1].emplace_back(v);
    R[r[v]].emplace_back(v);
  }
  dfs1(0);
  for (int i = 0; i < m; ++i) {
    if ((int)R[i].size() > kBuck) {
      isLarge[i] = true;
      ind[i] = large.size();
      large.emplace_back(i);
    }
  }
  dfs2(0);
  for (int r2 : large) {
    for (int v : R[r2]) {
      sum[in[v]] += 1;
    }
    for (int i = 1; i <= n; ++i) {
      sum[i] += sum[i - 1];
    }
    for (int v = 0; v < n; ++v) {
      root2[ind[r2]][r[v]] += sum[out[v]] - sum[in[v] - 1];
    }
    for (int i = 1; i <= n; ++i) {
      sum[i] = 0;
    }
  }
  for (int q = 0; q < Q; ++q) {
    int r1, r2;
    cin >> r1 >> r2;
    r1 -= 1, r2 -= 1;
    if ((int)R[r1].size() > kBuck) {
      cout << root1[ind[r1]][r2] << '\n';
    } else if ((int)R[r2].size() > kBuck) {
      cout << root2[ind[r2]][r1] << '\n';
    } else cout << solve(r1, r2) << '\n';
    cout.flush();
  }
  return 0;
}

• Subtask 1: 15 / 15
• Subtask 2: 85 / 85