제출 #1099687

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1099687		hengliao	수열 (APIO14_sequence)	C++14	100 / 100	274 ms	88008 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;
    }
    double xinter(const line &tar){
        return ((double) (tar.n-n)) / (m-tar.m);
    }
};
 
const ll mxN=1e5+5;
 
ll dp[mxN];
ll new_dp[mxN];
int par[mxN][201];
line dq[mxN];
ll fr, bk, timer;
bool good[mxN];
ll l[mxN];
ll r[mxN];
 
void solve(){
    memset(dp, 0, sizeof(dp));
    memset(par, 0, sizeof(par));
    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){
        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;
    }
    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[0]={0, 0, 0};
        memset(good, 0, sizeof(good));
        memset(l, -1, sizeof(l));
        memset(r, -1, sizeof(r));
        good[0]=1;
        fr=0;
        bk=0;
        timer=0;
        for(ll i=1;i<=siz;i++){
            while(fr!=bk && dq[fr].getval(pre[i])<dq[r[fr]].getval(pre[i])){
                fr=r[fr];
            }
            new_dp[i]=dq[fr].getval(pre[i]);
            par[i][tar]=dq[fr].id;
            line nline={pre[i], dp[i]-pre[i]*pre[i], i};
            while(fr!=bk && dq[bk].xinter(nline)<
                dq[l[bk]].xinter(nline)){
                bk=l[bk];
            }
            timer++; 
            r[bk]=timer;
            l[timer]=bk;
            dq[timer]=nline;
            bk=r[bk];
        }
        for(ll i=1;i<=siz;i++){
            dp[i]=new_dp[i];
        }
    }
    for(ll i=1;i<=siz;i++){
        dp[i]=dp[i]+(pre[siz]-pre[i])*pre[i];
    }
    ll ans=0;
    ll cur=1;
    for(ll i=1;i<=siz-1;i++){
        if(dp[i]>ans){
            ans=dp[i];
            cur=i;
        }
    }
    cout<<ans<<'\n';
    vll ans1;
    for(ll i=k;i>=1;i--){
        //assert(cur-1<-100);
        /*if(cur==0){
            cout<<"wrong\n";
            break;
        }*/
        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])


7 3
4 1 3 4 0 2 3
*/

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