Submission #646819

#TimeUsernameProblemLanguageResultExecution timeMemory
646819Alex_tz307Boxes with souvenirs (IOI15_boxes)C++17
100 / 100
471 ms196216 KiB
#include <bits/stdc++.h>
#include "boxes.h"
 
using namespace std;
 
const int64_t INF = 1e18L;
 
void minSelf(int64_t &x, int64_t y) {
  if (y < x) {
    x = y;
  }
}
 
long long delivery(int n, int k, int L, int p[]) {
  vector<int64_t> l(n + 1), r(n + 2);
  /// l[i] =def= costul minim sa livrez 1..i mergand doar la dreapta cand livrez si intorcandu-ma la inceput in ce directie vreau
  /// Analog r[i] in sens invers
  
  for (int i = 1; i <= n; ++i) {
    int last = max(0, i - k); /// mereu va fi optim sa aleg ultimii k si nu mai putin, pentru a parcurge o distanta mai mica inapoi inaintea lor, analog si in sens invers
    l[i] = l[last] + p[i - 1] + min(p[i - 1], L - p[i - 1]);
  }
 
  for (int i = n; i > 0; --i) {
    int nxt = min(n + 1, i + k);
    r[i] = r[nxt] + (L - p[i - 1]) + min(p[i - 1], L - p[i - 1]);
  }
 
  int64_t best = INF;
  for (int i = 0; i <= n; ++i) {
    minSelf(best, l[i] + r[i + 1]);
  }
 
  /// Observatie: Va fi maxim o tura de impartit cadouri in care parcurg cercul complet si aia va fi
  /// ultima, in care voi imparti toate cadourile ramase, asa ca fixez si variantele in care am k
  /// cadouri ramase si la final si parcurg tot cercul
  for (int i = 0; i <= n - k; ++i) {
    minSelf(best, l[i] + r[i + k + 1] + L);
  }
 
  return best;
}
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