제출 #678855

#제출 시각아이디문제언어결과실행 시간메모리
678855uroskFinancial Report (JOI21_financial)C++14
14 / 100
4065 ms16836 KiB
#define here cerr<<"===========================================\n"
#define dbg(x) cerr<<#x<<": "<<x<<endl;
#include "bits/stdc++.h"
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
#define ld double
#define ll long long
#define llinf 100000000000000000LL // 10^17
#define pb push_back
#define popb pop_back
#define fi first
#define sc second
#define endl '\n'
#define pll pair<ll,ll>
#define pld pair<ld,ld>
#define sz(a) (ll)(a.size())
#define all(a) a.begin(),a.end()
#define ceri(a,l,r) {cerr<<#a<<": ";for(ll i_ = l;i_<=r;i_++) cerr<<a[i_]<< " ";cerr<<endl;}
#define cer(a) {cerr<<#a<<": ";for(ll x_ : a) cerr<<x_<< " ";cerr<<endl;}

#define daj_mi_malo_vremena ios_base::sync_with_stdio(false);cerr.tie(0);cout.tie(0);cin.tie(0);

using namespace std;
//using namespace __gnu_pbds;
/*
ll add(ll x,ll y){
    x+=y;
    if(x<0){
        x%=mod;
        x+=mod;
    }else{
        if(x>=mod) x%=mod;
    }
    return x;
}
ll mul(ll a,ll b){
	ll ans = (a*b)%mod;
	if(ans<0) ans+=mod;
	return ans;
}
typedef tree<int,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef tree<int,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll rnd(ll l,ll r){
    return uniform_int_distribution<ll>(l,r)(rng);
}
*/
#define maxn 300005
ll n,d;
ll a[maxn];
ll t[2*maxn],t2[2*maxn],lazy[2*maxn],ls[2*maxn],rs[2*maxn],tsz = 0,root = 0;
void init(ll &v,ll tl,ll tr){
    v = ++tsz;
    if(tl==tr) return;
    ll mid = (tl+tr)/2;
    init(ls[v],tl,mid);
    init(rs[v],mid+1,tr);
}
void upd(ll v,ll tl,ll tr,ll i,ll x){
    if(tl==tr){t[v] = x;return;}
    ll mid = (tl+tr)/2;
    if(i<=mid) upd(ls[v],tl,mid,i,x);
    else upd(rs[v],mid+1,tr,i,x);
    t[v] = max(t[ls[v]],t[rs[v]]);
}
ll get(ll v,ll tl,ll tr,ll l,ll r){
    if(tr<l||tl>r||tl>tr) return 0;
    if(l<=tl&&tr<=r) return t[v];
    ll mid = (tl+tr)/2;
    return max(get(ls[v],tl,mid,l,r),get(rs[v],mid+1,tr,l,r));
}
void upd(ll v,ll x){
    t2[v] = x;
    lazy[v] = x;
}
void push(ll v){
    if(lazy[v]){
        upd(ls[v],lazy[v]);
        upd(rs[v],lazy[v]);
    }
    lazy[v] = 0;
}
void upd(ll v,ll tl,ll tr,ll l,ll r,ll x){
    if(r<tl||l>tr||tl>tr) return;
    if(l<=tl&&tr<=r){upd(v,x);return;}
    push(v);
    ll mid = (tl+tr)/2;
    upd(ls[v],tl,mid,l,r,x);
    upd(rs[v],mid+1,tr,l,r,x);
    t2[v] = min(t2[ls[v]],t2[rs[v]]);
}
ll get(ll v,ll tl,ll tr,ll i){
    if(tl==tr) return t2[v];
    push(v);
    ll mid = (tl+tr)/2;
    if(i<=mid) return get(ls[v],tl,mid,i);
    return get(rs[v],mid+1,tr,i);
}
void tc(){
    cin >> n >> d;
    for(ll i = 1;i<=n;i++) cin >> a[i];
    {
        set<ll> s;
        ll it = 0;
        map<ll,ll> mp;
        for(ll i = 1;i<=n;i++) s.insert(a[i]);
        for(ll x : s) mp[x] = ++it;
        for(ll i = 1;i<=n;i++) a[i] = mp[a[i]];
    }
    init(root,1,n);
    ll k = 1;
    for(ll i = 1;i<=n;i++){
        while(get(1,1,n,k)<i-d){
            upd(1,1,n,k,-llinf);
            k++;
        }
        ll x = a[i];
        ll y = max(1LL,get(root,1,n,1,x-1)+1);
        y = max(y,get(1,1,n,x,x));
        upd(1,1,n,x,y);
        for(ll j = max(x,k);j<=n;j++){
            ll z = max(y,get(1,1,n,j,j));
            upd(root,1,n,j,z);
        }
        upd(root,1,n,max(x,k),n,i);
    }
    cout<<t[root]<<endl;
}
int main(){
	daj_mi_malo_vremena
    int t; t = 1;
    while(t--){
        tc();
    }
	return 0;
}
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