제출 #467329

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
467329		ivan_tudor	휴가 (IOI14_holiday)	C++14	100 / 100	2179 ms	5752 KiB

holiday

#include"holiday.h"
#include <vector>
#include <iostream>
#include <set>
#include <algorithm>
using namespace std;
const int N = 100005;
vector<int> a;
multiset<int> big, small;
long long sum;
int canhold;
int L, R;
void insert_val(int v){
  big.insert(v);
  sum += v;
  while((int)big.size() > max(canhold, 0)){
    int lval = (*big.begin());
    sum -= lval;
    big.erase(big.find(lval));
    small.insert(lval);
  }
}
void erase_val(int v){
  if(small.find(v) != small.end()){
    small.erase(small.find(v));
  }
  else if(big.find(v) != big.end()){
    big.erase(big.find(v));
    sum -= v;
  }

  while((int)big.size() < canhold && small.size() > 0){
    int val = (*small.rbegin());
    big.insert(val);
    sum += val;
    small.erase(small.find(val));
  }
}
void transf(int newL, int newR){
  while(L > newL){
    canhold -=2;
    L--;
    insert_val(a[L]);
  }
  while(R < newR){
    canhold --;
    R++;
    insert_val(a[R]);
  }

  while(L < newL){
    canhold +=2;
    erase_val(a[L]);
    L++;
  }
  while(R > newR){
    canhold++;
    erase_val(a[R]);
    R--;
  }
}
long long ans = 0;
void solve(int l, int r, int opl, int opr){
  if(r < l)
    return;
  int mid = (l + r)/2;
  transf(mid, opl);
  long long maxi = sum;
  int  mpoz = opl;
  for(int i = opl; i <= opr; i++){
    transf(mid, i);
    if(maxi < sum){
      maxi = sum;
      mpoz = i;
    }
  }
  ans = max(ans, maxi);
  if(l == r)
    return;
  if(l <= mid - 1)
    solve(l, mid - 1, opl, mpoz);
  if(mid + 1 <= r)
    solve(mid + 1, r, mpoz, opr);
}
long long int findMaxAttraction(int n, int start, int d, int attraction[]) {
  a = vector<int>(n);
  for(int i = 0; i < n; i++)
    a[i] = attraction[i];

  big.clear();
  small.clear();
  big.insert(a[start]);
  sum = a[start];
  L = R = start;
  canhold = d;
  solve(0, start, start, n - 1);

  reverse(a.begin(), a.end());
  start = n - 1 - start;
  big.clear();
  small.clear();
  big.insert(a[start]);
  sum = a[start];
  L = R = start;
  canhold = d;
  solve(0, start, start, n - 1);
 // cerr<<op;
  return ans;
}

• Subtask 1: 7 / 7
• Subtask 2: 23 / 23
• Subtask 3: 17 / 17
• Subtask 4: 53 / 53