제출 #992033

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
992033		huutuan	휴가 (IOI14_holiday)	C++14	100 / 100	2463 ms	8020 KiB

holiday

#include"holiday.h"

#include <bits/stdc++.h>

using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;

#define int long long

template<class T, class U>
using ordered_set=tree<T, null_type, U, rb_tree_tag, tree_order_statistics_node_update>;

struct DS{
   int count;
   int sum;
   ordered_set<pair<int, int>, greater<pair<int, int>>> st;
   void clear(){
      // cerr << "clear" << endl;
      count=0;
      sum=0;
      st.clear();
   }
   DS (){
      clear();
   }
   int size(){
      return st.size();
   }
   void set_count(int new_count){
      new_count=max(new_count, 0ll);
      while (count<new_count){
         if (size()>count) sum+=st.find_by_order(count)->first;
         ++count;
      }
      while (count>new_count){
         --count;
         if (size()>count) sum-=st.find_by_order(count)->first;
      }
   }
   void insert(pair<int, int> x){
      // cerr << "insert " << x.second << endl;
      int id=st.order_of_key(x);
      if (id<count){
         sum+=x.first;
         st.insert(x);
         if (size()>count) sum-=st.find_by_order(count)->first;
      }else st.insert(x);
   }
   void erase(pair<int, int> x){
      // cerr << "erase " << x.second << endl;
      int id=st.order_of_key(x);
      if (id<count){
         sum-=x.first;
         st.erase(x);
         if (size()>=count) sum+=st.find_by_order(count-1)->first;
      }else st.erase(x);
   }
} st;

const int N=1e5;
int a[N], n, start, d, ans;

void dnc(int l, int r, int optl, int optr){
   if (l>r) return;
   int mid=(l+r)>>1;
   for (int i=mid; i<r; ++i) st.insert({a[i], i});
   st.set_count(d-(start-mid)-(optl-mid));
   int optmid=optl, optval=st.sum;
   for (int i=optl+1; i<=optr; ++i){
      st.set_count(d-(start-mid)-(i-mid));
      st.insert({a[i], i});
      if (optval<st.sum){
         optmid=i;
         optval=st.sum;
      }
   }
   ans=max(ans, optval);
   for (int i=mid; i<r; ++i) st.erase({a[i], i});
   for (int i=optl+1; i<=optr; ++i) st.erase({a[i], i});
   if (mid>0){
      // r -> optl -> mid - 1 -> optl
      for (int i=mid-1; i<r; ++i) st.insert({a[i], i});
      dnc(l, mid-1, optl, optmid);
      for (int i=mid-1; i<r; ++i) st.erase({a[i], i});
   }
   if (mid+1<n){
      // r -> optl -> r -> optmid
      for (int i=optl+1; i<=optmid; ++i) st.insert({a[i], i});
      // r -> optmid
      dnc(mid+1, r, optmid, optr);
      for (int i=optl+1; i<=optmid; ++i) st.erase({a[i], i});
   }
}

void solve(){
   for (int l=start; l>=0; --l){
      st.set_count(d-(start-l));
      st.insert({a[l], l});
      ans=max(ans, st.sum);
   }
   st.clear();
   st.insert({a[start], start});
   dnc(0, start, start, n-1);
   st.clear();
}

int findMaxAttraction(int32_t _n, int32_t _start, int32_t _d, int32_t _a[]) {
   n=_n; start=_start; d=_d;
   for (int i=0; i<n; ++i) a[i]=_a[i];
   solve();
   reverse(a, a+n);
   start=n-start-1;
   solve();
   return ans;
}

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