Submission #1300442

#	Time	Username	Problem	Language	Result	Execution time	Memory
1300442		rxlfd314	Sphinx's Riddle (IOI24_sphinx)	C++20	50 / 100	312 ms	1804 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);
  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) {
    int cur = N + 1;
    FOR(i, 0, N - 1)
      if (E[i] < 0 && uf.find(i) == i)
        E[i] = cur++;
    FOR(i, 0, N - 1)
      E[i] = E[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]);
    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;
      }
    }
    for (const auto &j : links)
      uf.unite(i, j);
  };
  dfs(dfs, 0);
  
  adj.assign(N, vt<int>());
  FOR(i, 0, M - 1) {
    adj[uf.find(X[i])].push_back(uf.find(Y[i]));
    adj[uf.find(Y[i])].push_back(uf.find(X[i]));
  }
  vt<int> A, B, seen(N);
  const auto colour = [&](auto &&self, const int i, const int c) -> void {
    (c ? A : B).push_back(i);
    seen[i] = 1;
    for (const auto &j : adj[i])
      if (!seen[j])
        self(self, j, !c);
  };
  colour(colour, uf.find(0), 0);
  
  vt<int> ans(N);
  FOR(_, 0, 1) {
    FOR(c, 0, N - 1) {
      vt<int> E(N);
      for (const auto &i : B)
        E[i] = c;
      int prv = -1;
      while (prv + 1 < size(A)) {
        FOR(i, 0, prv)
          E[A[i]] = N;
        FOR(i, prv + 1, size(A) - 1)
          E[A[i]] = -1;
        FOR(i, 0, N - 1)
          E[i] = E[uf.find(i)];
        if (query(E) == expect(E))
          break;
        int lo = prv + 1, hi = size(A) - 1;
        while (lo < hi) {
          const int mid = lo + hi >> 1;
          FOR(i, prv + 1, mid)
            E[A[i]] = -1;
          FOR(i, mid + 1, size(A) - 1)
            E[A[i]] = N;
          FOR(i, 0, N - 1)
            E[i] = E[uf.find(i)];
          if (query(E) == expect(E))
            lo = mid + 1;
          else
            hi = mid;
        }
        ans[A[lo]] = c;
        prv = lo;
      }
    }
    swap(A, B);
  }
  
  FOR(i, 0, N - 1)
    ans[i] = ans[uf.find(i)];
  return ans;
}

• Subtask 0: 0 / 0
• Subtask 1: 1.5 / 3
• Subtask 2: 3.5 / 7
• Subtask 3: 16.5 / 33
• Subtask 4: 10.5 / 21
• Subtask 5: 18 / 36