이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define deb(...) logger(#__VA_ARGS__, __VA_ARGS__)
template<typename ...Args>
void logger(string vars, Args&&... values) {
cerr<<vars<<" = ";
string delim="";
(...,(cerr<<delim<<values,delim=", "));
cerr<<"\n";
}
#else
#define deb(...) logger(#__VA_ARGS__, __VA_ARGS__)
template<typename ...Args>
void logger(string vars, Args&&... values) {}
#endif
#define pb push_back
#define sz(x) (int)(x.size())
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
template<typename T> inline void maxa(T& a,T b){a=max(a,b);}
template<typename T> inline void mina(T& a,T b){a=min(a,b);}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int mxN=4e5+5; //make sure this is right
const int mod=1e9+7;
struct RANK{
vector<int> all;
int get(int x){return lower_bound(all.begin(),all.end(),x)-all.begin()+1;}
void add(int x){all.pb(x);}
void fix(){
sort(all.begin(),all.end());
all.resize(unique(all.begin(),all.end())-all.begin());
}
} rnk;
struct TREE{
int st[4*mxN];
void update(int v,int l,int r,int ind,int val){
if(l>ind || r<ind) return;
if(l==r){
maxa(st[v],val);
} else{
int m=(l+r)>>1;
update(v<<1,l,m,ind,val);
update(v<<1|1,m+1,r,ind,val);
st[v]=max(st[v<<1],st[v<<1|1]);
}
}
int query(int v,int l,int r,int lq,int rq){
if(lq>rq)
return 0;
if(l>=lq && r<=rq){
return st[v];
}
int m=(l+r)>>1;
return max(query(v<<1,l,m,lq,min(m,rq)),query(v<<1|1,m+1,r,max(lq,m+1),rq));
}
} tree1,tree2;
int main(){
cin.sync_with_stdio(0);
cin.tie(0); cout.tie(0);
#ifdef LOCAL
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int n,x; cin>>n>>x;
vector<int> a(n),lis(n);
for(int &i:a){
cin>>i;
rnk.add(i);
rnk.add(i-x);
}
rnk.fix();
int ans=0;
int m=sz(rnk.all);
for(int i=n-1;i>=0;i--){
int lo=rnk.get(a[i]-x);
a[i]=rnk.get(a[i]);
int lis=tree1.query(1,1,m,a[i]+1,m);
int use=max(tree1.query(1,1,m,lo+1,m),tree2.query(1,1,m,a[i]+1,m));
maxa(ans,use+1);
tree1.update(1,1,m,a[i],lis+1);
tree2.update(1,1,m,a[i],use+1);
}
cout<<ans<<"\n";
}
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