Submission #1177610

#	Time	Username	Problem	Language	Result	Execution time	Memory
1177610		Kaztaev_Alisher	Sphinx's Riddle (IOI24_sphinx)	C++20	12 / 100	24 ms	5612 KiB

sphinx

#include <bits/stdc++.h>

#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define file(s) if (fopen(s".in", "r")) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)
#define all(a) a.begin() , a.end()
#define F first
#define S second

using namespace std;
using ll = long long;

const ll N = 2e5+5 , inf = 2e9 + 7;
const ll INF = 1e18 ,   mod = 1e9+7;

int perform_experiment(vector<int> E);
int n , was[N] , check[N] , p[N];
vector<int> g[N] , ord;
int gett(int x){
	if(p[x] == x) return x;
	return p[x] = gett(p[x]);
}
int merge(int a, int b){
	a = gett(a);
	b = gett(b);
	if(a == b) return 0 ;
	p[a] = b;
	return 1;
}
void dfs(int x) {
  was[x] = true;
  for (int i : g[x]) {
    if (!was[i] && ord[x] == ord[i]) dfs(i);
  }
}

int expected(int n) {
  for (int i = 0; i < n; i++) was[i] = false;

  int sum = 0;
  for (int i = 0; i < n; i++) {
    if (ord[i] == -1) {
      sum++;
    } else if (!was[i]) {
      sum++;
      dfs(i);
    }
  }

  return sum;
}
vector<int> find_colours(int N, vector<int> X, vector<int> Y) {
  vector<vector<int>> comps;
  comps.push_back({0});
  for (int i = 1; i < N; ++i) {
    vector<int> ord(N, -1);
    for (int j = i + 1; j < N; ++j) ord[j] = N;
    if (perform_experiment(ord) == (int)comps.size() + 1 + (i + 1 < N)) {
      comps.push_back({i});
      continue;
    }

    int lo = 0, hi = comps.size();
    while (lo + 1 < hi) {
      int mid = (lo + hi) / 2;
      ord.assign(N, N);
      ord[i] = -1;
      for (int c = mid; c < hi; ++c) {
        for (int u : comps[c]) {
          ord[u] = -1;
        }
      }
      int expected = 2 + hi - mid;
      if (perform_experiment(ord) < expected) {
        lo = mid;
      } else {
        hi = mid;
      }
    }
    comps[lo].push_back(i);
  }

  vector<int> F(N, -1);
  for (int c = 0; c < (int)comps.size(); ++c) {
    for (int u : comps[c]) {
      F[u] = c;
    }
  }
  return F;
}	

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