제출 #1300282

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1300282		rxlfd314	스핑크스 (IOI24_sphinx)	C++20	64 / 100	69 ms	652 KiB

sphinx

#include "sphinx.h"
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ari2 = array<int, 2>;
using ari3 = array<int, 3>;
using arl2 = array<ll, 2>;
using arl3 = array<ll, 3>;
template <class T> using vt = vector<T>;
#define all(x) begin(x), end(x)
#define size(x) (int((x).size()))
#define REP(a,b,c,d) for(int a=(b);(d)>0?a<=(c):a>=(c);a+=(d))
#define FOR(a,b,c) REP(a,b,c,1)
#define ROF(a,b,c) REP(a,b,c,-1)

struct DSU {
  vt<int> uf;
  DSU(const int n) : uf(n, -1) {}
  int find(const int i) { return uf[i] < 0 ? i : uf[i] = find(uf[i]); }
  bool unite(int a, int b) {
    if ((a = find(a)) == (b = find(b)))
      return false;
    if (uf[a] > uf[b])
      swap(a, b);
    uf[a] += uf[b];
    uf[b] = a;
    return true;
  }
};

int query(const vt<int> &E) { return perform_experiment(E); }

vt<int> find_colours(const int N, const vt<int> X, const vt<int> Y) {
  const int M = size(X);
  if (N <= 50) {
    vt<int> ret(N);
    FOR(i, 0, N-1)
      FOR(c, 0, N-1) {
        vt<int> E(N, c);
        E[i] = -1;
        if (query(E) == 1) {
          ret[i] = c;
          break;
        }
      }
    return ret;
  }
  if (M == N * (N - 1) / 2) {
    vt<int> ret(N);
    FOR(i, 0, N-1) {
      int lo = 0, hi = N-1;
      while (lo < hi) {
        const int mid = lo + hi >> 1;
        vt<int> E(N);
        E[i] = -1;
        int cur = 0;
        FOR(j, 0, N-1)
          if (j != i)
            E[j] = min(cur++, mid);
        if (query(E) > mid + 1)
          lo = mid + 1;
        else
          hi = mid;
      }
      ret[i] = lo;
    }
    return ret;
  }
  bool sub3 = true;
  FOR(i, 0, M - 1)
    sub3 &= X[i] == i && Y[i] == i + 1;
  if (sub3) {
    vt<int> ans(N);
    FOR(s, 0, 1) {
      vt<int> test, subject;
      REP(i, s, N - 1, 2)
        test.push_back(i);
      REP(i, !s, N - 1, 2)
        subject.push_back(i);
      FOR(c, 0, N - 1) {
        int prv = -1;
        while (prv + 1 < size(subject)) {
          vt<int> E(N);
          for (const auto &i : test)
            E[i] = c;
          FOR(i, 0, prv)
            E[subject[i]] = N;
          FOR(i, prv + 1, size(subject) - 1)
            E[subject[i]] = -1;
          if (query(E) == N)
            break;
          int lo = prv + 1, hi = size(subject) - 1;
          while (lo < hi) {
            const int mid = lo + hi >> 1;
            FOR(i, prv + 1, mid)
              E[subject[i]] = -1;
            FOR(i, mid + 1, size(subject) - 1)
              E[subject[i]] = N;
            if (query(E) == N)
              lo = mid + 1;
            else
              hi = mid;
          }
          ans[subject[lo]] = c;
          prv = lo;
        }
      }
    }
    return ans;
  }
  vt<vt<int>> adj(N);
  FOR(i, 0, M - 1) {
    adj[X[i]].push_back(Y[i]);
    adj[Y[i]].push_back(X[i]);
  }
  
  DSU uf(N);
  const auto expect = [&](vt<int> E) {
    FOR(i, 0, N - 1)
      if (E[i] < 0)
        E[i] = uf.find(i);
    DSU uf2(N);
    int ret = N;
    FOR(i, 0, M - 1)
      ret -= E[X[i]] == E[Y[i]] && uf2.unite(X[i], Y[i]);
    assert(ret == query(E));
    return ret;
  };
  
  int timer = 0;
  vt<int> tin(N, -1), tout(N), rev_tin(N);
  const auto dfs = [&](auto &&self, const int i) -> void {
    rev_tin[tin[i] = timer++] = i;
    vt<int> s_roots;
    for (const auto &j : adj[i])
      if (tin[j] < 0)
        s_roots.push_back(j), self(self, j);
    tout[i] = timer - 1;
    vt<int> links;
    for (const auto &j : s_roots) {
      vt<int> comps;
      FOR(k, tin[j], tout[j])
        comps.push_back(uf.find(rev_tin[k]));
      sort(all(comps));
      comps.erase(unique(all(comps)), comps.end());
      int prv = -1;
      while (prv + 1 < size(comps)) {
        vt<int> E(N, N);
        E[i] = -1;
        FOR(k, prv + 1, size(comps) - 1)
          E[comps[k]] = -1;
        FOR(k, 0, N - 1)
          E[k] = E[uf.find(k)];
        if (query(E) == expect(E))
          break;
        int lo = prv + 1, hi = size(comps) - 1;
        while (lo < hi) {
          const int mid = lo + hi >> 1;
          FOR(k, prv + 1, mid)
            E[comps[k]] = -1;
          FOR(k, mid + 1, size(comps) - 1)
            E[comps[k]] = N;
          FOR(k, 0, N - 1)
            E[k] = E[uf.find(k)];
          if (query(E) == expect(E))
            lo = mid + 1;
          else
            hi = mid;
        }
        links.push_back(comps[lo]);
        prv = lo + 1;
      }
    }
    for (const auto &j : links)
      uf.unite(i, j);
  };
  
  dfs(dfs, 0);
  vt<int> ans(N);
  FOR(i, 0, N - 1)
    ans[i] = uf.find(i);
  return ans;
}

• Subtask 0: 0 / 0
• Subtask 1: 3 / 3
• Subtask 2: 7 / 7
• Subtask 3: 33 / 33
• Subtask 4: 21 / 21
• Subtask 5: 0 / 36