Submission #521012

#	Time	Username	Problem	Language	Result	Execution time	Memory
521012		Wayne_Yan	Global Warming (CEOI18_glo)	C++17	27 / 100	113 ms	7628 KiB

glo

#include <bits/extc++.h>
using namespace std;
using namespace __gnu_pbds;
#define int long long
typedef int64_t ll;
typedef long double ld;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define pb emplace_back
#define mp make_pair
#define mt make_tuple
#define pii pair<int,int>
#define F(n) Fi(i,n)
#define Fi(i,n) Fl(i,0,n)
#define Fl(i,l,n) for(int i=l;i<n;i++)
#define RF(n) RFi(i,n)
#define RFi(i,n) RFl(i,0,n)
#define RFl(i,l,n) for(int i=n-1;i>=l;i--)
#define all(v) begin(v),end(v)
#define siz(v) (ll(v.size()))
#define get_pos(v,x) (lower_bound(all(v),x)-begin(v))
#define sort_uni(v) sort(begin(v),end(v)),v.erase(unique(begin(v),end(v)),end(v))
#define mem(v,x) memset(v,x,sizeof v)
#define ff first
#define ss second
#define mid ((l+r)>>1)
#define RAN(a,b) uniform_int_distribution<int> (a, b)(rng)
#define debug(x) (cerr << (#x) << " = " << x << "\n")
#define cmax(a,b) (a = max(a,b))
#define cmin(a,b) (a = min(a,b))

template <typename T> using max_heap = __gnu_pbds::priority_queue<T,less<T> >;
template <typename T> using min_heap = __gnu_pbds::priority_queue<T,greater<T> >;
template <typename T> using rbt = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

signed main(){
  
  int n,x;
  cin >> n >> x;
  int arr[n];
  int rev[n];
  F(n) cin >> arr[i];
  F(n) rev[i] = -arr[n-1-i];


  int ans = 0;

  pii suf[n]; // (length, greatest)
  
  vector<int> v;
  F(n){
    int c = get_pos(v, rev[i]);
    if(c == (int)v.size()){
      v.pb(rev[i]); 
    }else{
      cmin(v[c], rev[i]);
    }
    
    suf[n-1-i] = mp((int)v.size(), -v.back());
  }

  //F(n) printf("%d %d\n", suf[i].ff, suf[i].ss);

  v.clear();
  
  F(n){
    int c = get_pos(v, arr[i]);
    if( c == (int) v.size()){
      v.pb(arr[i]);
    }else{
      cmin(v[c], arr[i]);
    }

    if(i == n-1) break;
    int bruh = get_pos(v, suf[i+1].ss + x - 1);
    cmax(ans, suf[i+1].ff + bruh);
  }
  
  cmax(ans, (int)v.size());

  printf("%lld", ans);
  return 0;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 10
• Subtask 3: 0 / 13
• Subtask 4: 10 / 10
• Subtask 5: 0 / 20
• Subtask 6: 17 / 17
• Subtask 7: 0 / 25