제출 #949093

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
949093		hengliao	수열 (APIO14_sequence)	C++17	60 / 100	73 ms	131072 KiB

sequence

#include<bits/stdc++.h>
using namespace std;

#define F first
#define S second
#define pb push_back
#define vll vector<ll>
#define pll pair<ll, ll>
typedef long long ll;

struct line{
    ll m, n, id;
    ll getval(ll x){
        return m*x+n;
    }
    long double xinter(const line &tar){
        return (long double) (tar.n-n) / (long double) (m-tar.m);
    }
};

void solve(){
    ll n, k;
    cin>>n>>k;
    vll tep(n);
    for(ll i=0;i<n;i++){
        cin>>tep[i];
    }
    vll a;
    vll idx;
    for(ll i=0;i<n;i++){
        if(tep[i]!=0){
            a.pb(tep[i]);
            idx.pb(i);
        }
    }
    ll siz=a.size();
    if(siz<=k+1){
        ll ans=0;
        ll pre=0;
        for(ll i=0;i<siz;i++){
            ans+=a[i]*pre;
            pre+=a[i];
        }
        cout<<ans<<'\n';
        ll cnt=0;
        for(auto it:idx){
            if(it>=n-1) break;
            cout<<it+1<<' ';
            cnt++;
        }
        for(ll i=0;i<n && cnt<k;i++){
            if(tep[i]==0){
                cout<<i+1<<' ';
                cnt++;
            }
        }
        cout<<'\n';
        return;
    }
    vector<vll> dp(n+1, vll(k+1));
    vector<vll> par(n+1, vll(k+1));
    vll pre(siz+1);
    for(ll i=0;i<siz;i++){
        pre[i+1]=pre[i]+a[i];
    }
    //for()
    for(ll tar=2;tar<=k;tar++){
        deque<line> dq;
        dq.pb({0, 0, 0});
        for(ll i=1;i<=siz;i++){
            while(dq.size()>=2 && dq[0].getval(pre[i])<dq[1].getval(pre[i])){
                dq.pop_front();
            }
            dp[i][tar]=dq[0].getval(pre[i]);
            par[i][tar]=dq[0].id;
            line nline={pre[i], dp[i][tar-1]-pre[i]*pre[i], i};
            while(dq.size()>=2 && dq[(int) dq.size()-1].xinter(nline)<
                dq[(int) dq.size()-2].xinter(nline)){
                dq.pop_back();
            } 
            dq.pb(nline);
        }
    }
    for(ll i=1;i<=siz;i++){
        dp[i][k]=dp[i][k]+(pre[siz]-pre[i])*pre[i];
    }
    ll ans=0;
    ll cur=1;
    for(ll i=1;i<=siz;i++){
        if(dp[i][k]>ans){
            ans=dp[i][k];
            cur=i;
        }
    }
    cout<<ans<<'\n';
    vll ans1;
    for(ll i=k;i>=1;i--){
        //assert(cur-1<-100);
        ans1.pb(idx[cur-1]+1);
        //cout<<cur<<' ';
        cur=par[cur][i];
    }
    reverse(ans1.begin(), ans1.end());
    for(auto it:ans1){
        cout<<it<<' ';
    }
    cout<<'\n';
}

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    solve();
    return 0;
}

/*

dp[k][i]=max j<i dp[k-1][j]+(pre[i]-pre[j])*pre[j]
                =pre[j]*pre[i]+(dp[k-1][j]-pre[j]*pre[j])

*/

• Subtask 1: 0 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 17 / 17
• Subtask 5: 21 / 21
• Subtask 6: 0 / 29