Submission #421030

#	Time	Username	Problem	Language	Result	Execution time	Memory
421030		davi_bart	Financial Report (JOI21_financial)	C++17	48 / 100	4049 ms	11504 KiB

Main

#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define int ll
#define fi first
#define se second
#define ld long double
#define pb push_back
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int N,D;
vector<int> v;
int memo[300010];
int val[300010];
int prev[300010];
const int dim=100;
vector<pair<int,int> > ma(100000);
struct min_queue{
    deque<pair<int,int> >q;
    int top(){
        return q.front().fi;
    }
    void insert(int v,int t){
        while(q.size() && q.back().fi>=v)q.pop_back();
        q.pb({v,t});
    }
    void pop(int t){
        while(q.size() && q.front().se<=t)q.pop_front();
    }
}q;
signed main(){
	ios::sync_with_stdio(false);cin.tie(0);
    cin>>N>>D;
    vector<int> p={0};
    v.pb(0);
    for(int i=0;i<N;i++){
        int a;cin>>a;
        v.pb(a);
        p.pb(a);
    }
    /*sort(p.begin(),p.end());
    p.resize(unique(p.begin(),p.end())-p.begin());
    map<int,int> l;
    for(int i=0;i<p.size();i++)l[p[i]]=i;
    for(int i=0;i<=N;i++)v[i]=l[v[i]];*/


    for(int i=0;i<=N;i++){
        q.insert(v[i],i);
        q.pop(i-D);
        val[i]=q.top();
    }
    int ans=0;
    for(int i=1;i<=N;i++){
        memo[i]=1;
        int pos=i-i%dim;
        for(int j=pos;j>=0;j-=dim){
            if(ma[j/dim].fi<v[i]){
                for(int h=j;h<min(i,j+dim);h++){
                    if(v[h]<v[i]){
                        memo[i]=max(memo[i],memo[h]+1);
                    }
                }
            }
            else{
                for(int h=min(i-1,j+dim-1);h>=j;h--){
                    if(val[h]>=v[i])break;
                    if(v[h]<v[i]){
                        memo[i]=max(memo[i],memo[h]+1);
                    }
                }
                break;
            }
        }

        ans=max(ans,memo[i]);
        ma[i/dim].se=max(ma[i/dim].se,memo[i]);
        ma[i/dim].fi=max(ma[i/dim].fi,val[i]);
    }
    cout<<ans<<endl;

}

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 20 / 20
• Subtask 4: 0 / 12
• Subtask 5: 0 / 5
• Subtask 6: 0 / 35