Submission #605220

#	Time	Username	Problem	Language	Result	Execution time	Memory
605220		Evirir	Global Warming (CEOI18_glo)	C++17	100 / 100	88 ms	10080 KiB

glo

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
//using namespace __gnu_pbds;

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")
#define watch(x) cout<<(#x)<<"="<<(x)<<'\n'
#define mset(d,val) memset(d,val,sizeof(d))
#define cbug if(DEBUG) cout
#define setp(x) cout<<fixed<<setprecision(x)
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define forn(i,a,b) for(int i=(a);i<(b);i++)
#define fore(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
#define F first
#define S second
#define fbo find_by_order
#define ook order_of_key
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> ii;
typedef vector<ll> vi;
typedef vector<ii> vii;
//template<typename T>
//using pbds = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
void SD(int t=0){ cout<<"PASSED "<<t<<endl; }
ostream& operator<<(ostream &out, ii x){ out<<"("<<x.F<<","<<x.S<<")"; return out; }
template<typename T> void amax(T &a, T b){ a=max(a,b); }
template<typename T> void amin(T &a, T b){ a=min(a,b); }
const ll INF = ll(1e18);
const int MOD = 998244353;

const bool DEBUG = 0;
const int MAXN = 200005;
const int LG = 21;

int n;
ll X;
ll a[MAXN];
vi dpl, dpr;
vii chg;

int main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	cin>>n>>X;
	forn(i,0,n) cin>>a[i];

	dpl.resize(n + 1, INF);
	dpl[0] = -INF;
	dpr.resize(n + 1, -INF);
	dpr[0] = INF;
	chg.resize(n);

	for (int i = n - 1; i >= 0; i--)
	{
		int p = lower_bound(all(dpr), a[i], greater<ll>()) - dpr.begin();
		chg[i] = {p, dpr[p]};
		dpr[p] = a[i];
	}

	int ans = 0;
	forn(i, 0, n)
	{
		int pl = lower_bound(all(dpl), a[i]) - dpl.begin();
		int pr = chg[i].F;
		dpl[pl] = a[i];
		dpr[pr] = chg[i].S;

		int q = lower_bound(all(dpr), dpl[pl] - X, greater<ll>()) - dpr.begin() - 1;
		ans = max(ans, pl + q);

		q = lower_bound(all(dpl), dpr[pr] + X) - dpl.begin() - 1;
		ans = max(ans, q + pr);
	}

	int cur = 0;
	while (cur + 1 <= n && dpl[cur + 1] != INF)
		cur++;
	ans = max(ans, cur);

	cout << ans << '\n';
	
	return 0;
}

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