Submission #424862

#	Time	Username	Problem	Language	Result	Execution time	Memory
424862		ollel	Boxes with souvenirs (IOI15_boxes)	C++14	10 / 100	1 ms	304 KiB

boxes

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

#define rep(i,a,b) for(int i = a; i < b; i++)
#define pb push_back
#define lso(x) x&(-x)

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;

int n, k, l;
vi p;
vi pos;

ll min(ll a, int b) {return min(a, (ll)b);}

ll kone() {
  ll ans = 0;
  rep(i,0,n) {
    ans += 2*min(pos[i], l - pos[i]);
  }
  return ans;
}

ll kn() {
  ll ans = l;
  ans = min(ans, 2*(l - pos[0]));
  ans = min(ans, 2*pos[n - 1]);
  rep(i,0,n - 1) {
    ans = min(ans, 2*pos[i] + 2*(l - pos[i + 1]));
  }
  return ans;
}

ll solve(vi& ps) {
  int a = 0, b = min(k - 1, n - 1);
  ll ans = 0;
  int fhc = l / 2;
  int shc = l / 2 + (l & 1);

  while (a < n) {
    if (ps[b] <= fhc) ans += 2*ps[b];
    else if (ps[a] >= shc) ans += 2*(l - ps[a]);
    else {
      int low = a, high = b + 1, c;
      while (high - low > 1) {
        c = (high + low) / 2;
        if (ps[c] >= shc) high = c;
        else low = c;
      }
      ll cand1 = n, cand2 = 2*ps[c] + 2*(l - ps[c + 1]);
      ans += min(cand1, cand2);
    }

    a = b + 1;
    b = min(a + k - 1, n - 1);

  }
  return ans;
}

ll get() {
  ll ans = solve(pos);
  vi ppp(n);
  rep(i,0,n) ppp[i] = abs(l - pos[n - 1 - i]);
  ans = min(ans, solve(ppp));
  return ans;
}

ll delivery(int N, int K, int L, int positions[]) {
  n = N; k = K; l = L;
  pos.resize(n); rep(i,0,n) pos[i] = positions[i];
  return get();
  if (k == 1) return kone();
  else if (k == n) return kn();
  else return get();
}


// int main() {
//   int a = 7, b = 2, c = 8;
//   int pp[7] = {1, 1, 1, 1, 5, 6, 7};
//   cout << delivery(a, b, c, pp);
// }

Compilation message (stderr)

boxes.cpp: In function 'll solve(vi&)':
boxes.cpp:54:35: warning: 'c' may be used uninitialized in this function [-Wmaybe-uninitialized]
   54 |       ll cand1 = n, cand2 = 2*ps[c] + 2*(l - ps[c + 1]);
      |                                   ^

• Subtask 1: 10 / 10
• Subtask 2: 0 / 10
• Subtask 3: 0 / 15
• Subtask 4: 0 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 30