제출 #220098

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
220098		rama_pang	Minerals (JOI19_minerals)	C++14	100 / 100	87 ms	4260 KiB

minerals

#include "minerals.h"
#include <bits/stdc++.h>
using namespace std;

int optimal_pivot(int n) {
  static int MAXN = 43005;
  static vector<pair<int, int>> optimal(MAXN, make_pair(1e8, 1e8));
  static bool computed = false;

  if (computed) return optimal[n].second;

  computed = true;
  optimal[0] = optimal[1] = {0, 0};
  for (int i = 2; i < MAXN; i++) {
    int lo = 1, hi = i / 2;
    while (lo < hi) {
      int m1 = (lo + hi) / 2;
      int m2 = m1 + 1;
      int c1 = optimal[m1].first + optimal[i - m1].first + i - 1 + m1;
      int c2 = optimal[m2].first + optimal[i - m2].first + i - 1 + m2;
      if (c1 < c2) {
        hi = m1;
      } else {
        lo = m2;
      }
    }
    optimal[i] = make_pair(optimal[lo].first + optimal[i - lo].first + lo + i - 1, lo);
  }

  return optimal_pivot(n);
}

bool query(int x) {
  static int last = -1;
  int cur = Query(x);
  bool res = (cur == last);
  last = cur;
  return res;
}

void Solve(vector<int> A, vector<int> B, bool isBInside) {
  int n = A.size();
  if (n == 0) return;
  if (n == 1) return Answer(A[0], B[0]);

  int piv = optimal_pivot(n);
  vector<int> Aleft, Aright;
  vector<int> Bleft, Bright;
  if (isBInside) { // All of B is in the box, so we remove the second half
    for (int i = 0; i < n - piv; i++) {
      Bleft.emplace_back(B[i]);
    }
    for (int i = n - piv; i < n; i++) {
      Bright.emplace_back(B[i]);
      query(B[i]); // 
    }
  } else { // None of B is in the box, so we add the first half
    for (int i = 0; i < piv; i++) {
      Bleft.emplace_back(B[i]);
      query(B[i]); // add elemens in B such that only the first half remains
    }
    for (int i = piv; i < n; i++) {
      Bright.emplace_back(B[i]);
    }
  }

  for (int i = 0; i < n; i++) {
    if (Aright.size() == Bright.size()) {  
      Aleft.emplace_back(A[i]);   // all A[i] in the right half is matched
    } else if (Aleft.size() == Bleft.size()) { 
      Aright.emplace_back(A[i]);  // all A[i] in the left half is matched
    } else if (query(A[i])) {
      Aleft.emplace_back(A[i]);   // A[i] has a match is in the first half of B
    } else {
      Aright.emplace_back(A[i]);  // A[i] has a match in the second half of B
    }
  }

  return Solve(Aleft, Bleft, true), Solve(Aright, Bright, false);
}

void Solve(int N) {
  vector<int> A, B;
  for (int i = 1; i <= 2 * N; i++) {
    if (query(i)) { // the current element is already in the box
      B.emplace_back(i);
    } else { // the current element is a new element
      A.emplace_back(i);
    }
  }

  return Solve(A, B, true);
}

• Subtask 1: 6 / 6
• Subtask 2: 25 / 25
• Subtask 3: 9 / 9
• Subtask 4: 30 / 30
• Subtask 5: 5 / 5
• Subtask 6: 5 / 5
• Subtask 7: 5 / 5
• Subtask 8: 5 / 5
• Subtask 9: 10 / 10