제출 #422650

#제출 시각아이디문제언어결과실행 시간메모리
422650KalasLavasGlobal Warming (CEOI18_glo)C++14
100 / 100
75 ms5828 KiB
//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma,sse,sse2")
#include <bits/stdc++.h>
using namespace std;

mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
//#undef LOCALKL
#define IO \
ios_base::sync_with_stdio(0);(cin).tie(0);(cout).tie(0)
#define y1 y1_
#define prev prev_
#define all(a) (a).begin(),(a).end()
#define pb push_back
#define eb emplace_back
#define F first
#define S second
#ifdef LOCALKL
// #define cerr cerr<<"\33[1;32m"
// #define cout cout<<"\33[0m"
#else
#ifndef OO
#define endl '\n'
#define cerr if(1){}else cerr
#endif
#endif
#define OK cout<<"OK\n"<<endl;
#define setpre(k) fixed<<setprecision(k)
#define mmset(k,y) memset(k,y,sizeof(k))
#define sz(x) ((int)(x).size())
using pii = pair<int,int>;
using pll = pair<long long,long long>;
using ull = unsigned long long;
using intt = long long;
using ll = long long;
using ld = long double;

const ll m9 = 998244353;
const ll m7 = 1000000007;
const ll m18 = 1000000000000000000;
const ll i127 = 2139062143;
const ll l127 = 9187201950435737471;

int n, d, a[200002], lis[200002], lds[200002], ans, x;
int l, r, mid;
stack<pii>h;
int main()
{
    IO;
    memset(lis, 127, sizeof(lis));
    lis[0]=0;
    lds[0]=i127;
    cin>>n>>d;
    for(int i=1;i<=n;i++) cin>>a[i];

   	for(int i=1;i<=n;i++)
   	{
   		l=0; r=n;
   		while(l<r)
   		{
   			mid=(l+r)>>1;
   			if(lis[mid]<a[i]) l=mid+1;
   			else r=mid;
   		}
   		h.emplace(l, lis[l]);
   		lis[l]=a[i];
   		ans=max(ans, l);
   		// cerr<<l<<' '<<h.top().F<<' '<<h.top().S<<endl;
   	}
   	for(int i=n;i>=1;i--)
   	{
   		lis[h.top().F] = h.top().S;
   		h.pop();
   		l=0; r=n;
   		a[i]+=d;
   		while(l<r)
   		{
   			mid=(l+r)>>1;
   			if(a[i]<lds[mid]) l=mid+1;
   			else r=mid;
   		}
   		lds[l] = a[i];
   		x = l;
   		l=0; r=n;
   		while(l<r)
   		{
   			mid=(l+r)>>1;
   			if(lis[mid]<a[i]) l=mid+1;
   			else r=mid;
   		}
   		l--;
   		ans=max(ans, x+l);
   		// cerr<<i<<' '<<a[i]<<" : "<<l<<' '<<lis[l]<<" -> "<<x<<' '<<lds[x]<<endl;

   	}
   	cout<<ans<<endl;
    return 0;
}
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