Submission #958117

#TimeUsernameProblemLanguageResultExecution timeMemory
958117zeta7532Sequence (APIO23_sequence)C++17
40 / 100
2073 ms57168 KiB
#include "sequence.h"
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = int;
const ll mod = 998244353;
#define fi first
#define se second
#define rep(i,n) for(ll i=0;i<n;i++)
#define all(x) x.begin(),x.end()
#define faster ios::sync_with_stdio(false);cin.tie(nullptr)
 
pair<int,int> f(pair<int,int> a,pair<int,int> b){
    return {min(a.fi,b.fi),max(a.se,b.se)};
}
 
template<typename T>
struct RMQ{
    const T INF={1LL<<30,-1LL<<30};
    ll N;
    vector<T> dat;
    RMQ(ll N_):N(),dat(N_*4,INF){
        ll x=1;
        while(N_>x) x*=2;
        N=x;
    }
    void update(ll i,int x){
        i+=N-1;
        dat[i]={x,x};
        while(i>0){
            i=(i-1)/2;
            dat[i]=f(dat[i*2+1],dat[i*2+2]);
        }
    }
    T query(ll a,ll b){return query_sub(a,b,0,0,N);}
    T query_sub(ll a,ll b,ll k,ll l,ll r){
        if(r<=a||b<=l){
            return INF;
        }else if(a<=l&&r<=b){
            return dat[k];
        }else{
            T vl=query_sub(a,b,k*2+1,l,(l+r)/2);
            T vr=query_sub(a,b,k*2+2,(l+r)/2,r);
            return f(vl,vr);
        }
    }
};
 
int f(int N,vector<int> A){
    vector<int> B=A;
    sort(all(B));
    B.erase(unique(all(B)),B.end());
    vector<int> C(N,1);
    vector<vector<int>> G(N+1);
    rep(i,N) G[A[i]].push_back(i);
    int ans=0;
    rep(i,(int)B.size()){
        RMQ<pair<int,int>> seg(N+1);
        vector<int> cum(N+1,0);
        rep(j,N) cum[j+1]=cum[j]+C[j];
        rep(j,N+1) seg.update(j,cum[j]);
        int l=0,r=0;
        while(1){
            if(r+1==(int)G[B[i]].size()) break;
            auto e1=seg.query(0,G[B[i]][l]+1);
            auto e2=seg.query(G[B[i]][r+1]+1,N+1);
            if(e1.fi<=e2.se&&e2.fi<=e1.se){
                r++;
            }else{
                l++;
            }    
            if(r<l) r++;
            ans=max(ans,r-l+1);
        }
        for(int j:G[B[i]]){
            C[j]=-1;
        }
    }
    return ans;
}
 
int sequence(int N, std::vector<int> A) {
    int ans=0;
    vector<int> B=A;
    sort(all(B));
    rep(j,2){
        int cnt=0;
        rep(i,N) if(A[i]==B[(N-1+j)/2]) cnt++;
        ans=max(ans,cnt);
    }
    ans=max(ans,f(N,A));
    rep(i,N) A[i]=N+1-A[i];
    ans=max(ans,f(N,A));
    return ans;
}
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