Submission #1258162

#	Time	Username	Problem	Language	Result	Execution time	Memory
1258162		madamadam3	3개의 봉우리 (IOI25_triples)	C++20	75.09 / 100	1258 ms	40220 KiB

triples

#include "triples.h"
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using pi = pair<int, int>;

#define FOR(i, a, b) for (int i = a; i < b; i++)
#define ROF(i, a, b) for (int i = a; i >= b; i--)
#define each(a, x) for (auto &a : x)
#define all(x) (x).begin(), (x).end()
#define bg(x) (x).begin()
#define en(x) (x).end()
#define rev(x) reverse(all(x))
#define sz(x) int((x).size())
#define srt(x) sort(all(x))
#define cnt(a, x) count(all(x), a)
#define trace(x) each(a, (x)) cout << a << " "
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound

ll brute(vi h) {
  int n = sz(h);
  ll ans = 0;

  FOR(i, 0, n) {
    FOR(j, i+1, n) {
      FOR(k, j+1, n) {
        int d1 = min({k-j, k-i,j-i}), d3 = max({k-j, j-i, k-i}), f1 = min({h[i], h[j], h[k]}), f3 = max({h[i], h[j], h[k]}); 
        int d2 = (k-j) ^ (k-i) ^ (j-i) ^ d1 ^ d3, f2 = h[i] ^ h[j] ^ h[k] ^ f1 ^ f3;
        if (d1 == f1 && d2 == f2 && d3 == f3) ans++;
      }
    }
  }

  return ans;
}

struct Triple {
  int i, j, k;

  Triple() {};
  Triple(int I, int J, int K) {
    i = min({I, J, K});
    k = max({I, J, K});
    j = (I ^ J ^ K) ^ i ^ k; // lol
  }

  const bool operator<(const Triple &other) const {
    if (i != other.i) return i < other.i;
    if (j != other.j) return j < other.j;
    
    return k < other.k;
  }
};

ll count_triples(vi h) {
  int n = sz(h);

  ll ans = 0;
  FOR(i, 0, n) { // count number of triples where h[i] = d(i, j) or d(i, k)
    int H = h[i]; 
    set<Triple> trips; // prevent double count with dupe height

    int j = i + H;
    if (j < n) {
      int k1 = i + h[j], k2 = j + h[j]; // 2 cases — h[j] = d(i, k) or h[j] = d(j, k)
      if (k1 < n && h[k1] == (k1 - j)) trips.insert(Triple(i, j, k1)); // we know d(i, j) and d(i, k) to be h[i] and h[j], so does h[k] == d(j, k)
      if (k2 < n && h[k2] == (k2 - i)) trips.insert(Triple(i, j, k2)); // symmetric but this time for i
    }

    int k = i + H;
    if (k < n) {
      int j1 = i + h[k], j2 = k - h[k]; // h[k] = d(i, k), h[k] = d(j, k)
      if ((i < j1 && j1 < k) && h[j1] == (k - j1)) trips.insert(Triple(i, j1, k));
      if ((i < j2 && j2 < k) && h[j2] == (j2 - i)) trips.insert(Triple(i, j2, k));
    }

    for (auto &el : trips) {
      if (h[el.j] != h[el.i] && h[el.k] != h[el.i]) ans++;
    }
  }

  unordered_map<int, vector<int>> pc, nc; // positive line intercepts, negative line intercepts
  vi inp(n), inn(n); // lookup for the intercept
  FOR(i, 0, n) {
    int c1 = h[i] - i, c2 = h[i] + i;
    inp[i] = c1, inn[i] = c2;
    pc[c1].pb(i), nc[c2].pb(i);
  }

  for (auto &cnst : pc) { // count number of triples where h[j] = d(i, j), h[k] = d(i, k), and h[i] = d(j, k) = h[k] - h[j]
    int i = -cnst.first;
    if (i < 0 || i >= n) continue;

    int H = h[i];
    for (auto &el : cnst.second) {
      int j = el, k = el + H;

      if (k >= n) break;
      if (inp[k] != inp[j]) continue;

      ans++;
    }
  }

  FOR(j, 0, n) { // count number of triples where h[j] = d(i, k), h[k] = d(i, j) and h[i] = d(j, k) = h[j] - h[k]
    int f1 = lower_bound(all(pc[inp[j]]), j-h[j]) - bg(pc[inp[j]]);
    int f2 = upper_bound(all(nc[inn[j]]), j+h[j]) - bg(nc[inn[j]]);

    int p1 = lower_bound(all(pc[inp[j]]), j) - bg(pc[inp[j]]), p2 = lower_bound(all(nc[inn[j]]), j) - bg(nc[inn[j]]);
    if (p1 - f1 < f2 - p2) { // there is no way to set up the diagonals such that no matter which way you go its n^2
      for (int el = f1; el < p1; el++) {
        int i = pc[inp[j]][el];
        int dist = j - i;
        if (j + h[i] >= n || h[j + h[i]] != dist) continue;
        ans++;
      }
    } else {
      for (int el = p2 + 1; el < f2; el++) {
        int k = nc[inn[j]][el];
        int dist = k - j;
        if (j - h[k] < 0 || h[j - h[k]] != dist) continue;
        ans++;
      }
    }
  }

  return ans;
}

bool accept(double cur_score, double next_score, double temp, mt19937 &rng) {
  if (next_score > cur_score) return true;

  cur_score = -cur_score; next_score = -next_score;
  double prob = exp(-(next_score - cur_score) / temp);
  return generate_canonical<double, 53>(rng) < prob;
}

vi construct_range(int M, int K) {
  vi best(M, 1);

  int score = 0, best_score = 0;
  for (int tl = 0; tl < 2; tl++) {
  vi cur(M, 1);
  score = 0;

  // double T = 10000, u = 0.9995;
  double t0 = 100, c = 1.0;
  int trials = 1;

  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  uniform_int_distribution u(0, M-1);
  geometric_distribution<int> g(0.4);
  double lambda = 0.1; vector<double> w(M); for (int i = 0; i < M; i++) w[i] = exp(-lambda * i);
  discrete_distribution<int> d(all(w));

  // while (T > 1) {
  int lim = (150'000 / M) * 20;
  while(trials < lim && best_score < K) {
    if (best_score >= K) break;
    if (trials % 10000 == 0) cerr << "Trials: " << trials << " Best: " << best_score << "\n";

    double T = c / log(t0 + double(trials));

    vi nxt = cur;
    int idx = u(rng);
    // int val = g(rng) + 1; while (val >= M) val = g(rng) + 1;
    int val = u(rng); while (!(1 <= val && val < M)) val = u(rng);

    nxt[idx] = val;

    int ns = count_triples(nxt);
    if (accept(score, ns, T, rng)) {
      score = ns; cur = nxt;

      if (ns > best_score) {
        best_score = ns; best = nxt;
      }
    }

    // T *= u;
    trials++;
  }

  }
  cerr << "Array has: " << count_triples(best) << " triples.\n";
  return best;
}

• Subtask 0: 0 / 0
• Subtask 1: 8 / 8
• Subtask 2: 6 / 6
• Subtask 3: 10 / 10
• Subtask 4: 11 / 11
• Subtask 5: 16 / 16
• Subtask 6: 19 / 19
• Subtask 7: 5 / 5
• Subtask 8: 0.0925 / 5
• Subtask 9: 0 / 5
• Subtask 10: 0 / 5
• Subtask 11: 0 / 5
• Subtask 12: 0 / 5