제출 #1113371

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1113371		VinhLuu	Line Town (CCO23_day1problem3)	C++17	6 / 25	2076 ms	65760 KiB

Main

#include <bits/stdc++.h>
#define int long long
#define ll long long
#define all(lpv) lpv.begin(), lpv.end()
#define fi first
#define se second
using namespace std;

typedef pair<int,int> pii;
const int N = 1e6 + 5;
const int oo = 1e16;
const int mod = 1e6 + 3;

int n, a[N], m, zero;

void add(int &x,int y){
  if(x == -1) x = y;
  else x = min(x, y);
}

int bit[N];

void update(int i,int val){
  for(; i <= n; i += i & -i) bit[i] += val;
}

int get(int i){
  int ret = 0;
  for(; i; i -= i & -i) ret += bit[i];
  return ret;
}

namespace brute{
  vector<int> even[N], odd[N];
  void solve(){
    int ans = -1;
    for(int k = 0; k <= n - zero; k ++){
      for(int i = 0; i <= m; i ++){
        even[i].clear();
        odd[i].clear();
      }
      for(int i = n; i >= 1; i --){
        if(a[i] == 0){
          even[0].push_back(i);
          continue;
        }
        int x = abs(a[i]);
        if(i & 1){
          if(a[i] > 0) even[x].push_back(i);
          else odd[x].push_back(i);
        }else{
          if(a[i] > 0) odd[x].push_back(i);
          else even[x].push_back(i);
        }
      }
//      cerr << k << " start\n";
      bool ff = true;
      int cnt = 0;
      for(int i = 1; i <= n; i ++) bit[i] = 0;
      for(int i = 1; i <= k; i ++){
        int x = 1;
        if(i & 1){
          if(odd[x].empty()){
            ff = false;
            break;
          }
          int pos = odd[x].back();
          odd[x].pop_back();
          cnt += pos - 1 - get(pos - 1);
          update(pos, 1);
        }else{
          if(even[x].empty()){
            ff = false;
            break;
          }
          int pos = even[x].back();
          even[x].pop_back();
          cnt += pos - 1 - get(pos - 1);
          update(pos, 1);
        }
      }
//      cerr << ff << " " << cnt << " g\n";
      if(!ff) continue;
      while(!even[0].empty()){
        int pos = even[0].back();
        even[0].pop_back();
        cnt += pos - 1 - get(pos - 1);
        update(pos, 1);
      }
//      cerr << k << " " << cnt << " y\n";
      for(int i = k + zero + 1; i <= n; i ++){
        int x = 1;
        if(i & 1){
          if(even[x].empty()){
            ff = false;
            break;
          }
          int pos = even[x].back();
          even[x].pop_back();
          cnt += pos - 1 - get(pos - 1);
          update(pos, 1);
        }else{
          if(odd[x].empty()){
            ff = false;
            break;
          }
          int pos = odd[x].back();
          odd[x].pop_back();
          cnt += pos - 1 - get(pos - 1);
          update(pos, 1);
        }
      }
      if(!ff) continue;
//      cerr << k << " " << cnt << " result\n";
      add(ans, cnt);
    }
    cout << ans;
  }
}

signed main(){
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  #define task "v"
  if(fopen(task ".inp","r")){
    freopen(task ".inp","r",stdin);
    freopen(task ".out","w",stdout);
  }
  cin >> n;
  vector<int> rrh;
  for(int i = 1; i <= n; i ++){
    cin >> a[i];
    zero += (a[i] == 0);
    if(a[i] != 0) rrh.push_back(abs(a[i]));
  }
  sort(all(rrh)); rrh.resize(unique(all(rrh)) - rrh.begin());
  m = (int)rrh.size();
  for(int i = 1; i <= n; i ++){
    if(a[i] == 0) continue;
    if(a[i] < 0) a[i] = -(lower_bound(all(rrh), abs(a[i])) - rrh.begin() + 1);
    else a[i] = lower_bound(all(rrh), abs(a[i])) - rrh.begin() + 1;
  }
  brute :: solve();
}
/*
5
0 -1 -1 0 1
*/

컴파일 시 표준 에러 (stderr) 메시지

Main.cpp: In function 'int main()':
Main.cpp:125:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  125 |     freopen(task ".inp","r",stdin);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~
Main.cpp:126:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  126 |     freopen(task ".out","w",stdout);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 3 / 3
• Subtask 2: 0 / 3
• Subtask 3: 3 / 3
• Subtask 4: 0 / 4
• Subtask 5: 0 / 4
• Subtask 6: 0 / 3
• Subtask 7: 0 / 2
• Subtask 8: 0 / 3