Submission #1095639

#	Time	Username	Problem	Language	Result	Execution time	Memory
1095639		akim9905	Split the sequence (APIO14_sequence)	C++17	71 / 100	2047 ms	37488 KiB

sequence

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

#define fileio() freopen("input.txt","r",stdin); freopen("output.txt","w",stdout)
#define fio() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define all(x) (x).begin(), (x).end()
#define allr(x) x.rbegin(), x.rend()
#define cmprs(x) sort(all(x)),x.erase(unique(all(x)),x.end())
#define endl "\n"
#define sp " "
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define F first
#define S second
#define rz resize
#define sz(a) (int)(a.size())
#define clr(a) a.clear()

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<int, ll> pil;
typedef tuple<int, int, int> tpi;
typedef tuple<ll, ll, ll> tpl;
typedef pair<double, ll> pdl;
typedef pair<double, int> pdi;

const int dx[] = {1,-1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
const ll MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 101010; // PLZ CHK!
const int MAXM = 202;

struct LiChao {
    struct Line {
        ll a,b,i;
        pll get(ll x) {return {a*x+b,i};}
    };

    struct Node {
        int l,r;
        ll s,e;
        Line line;
    };

    vector<Node> tree;
    void init(ll s, ll e) { tree.pb({-1,-1,s,e,{0,-LINF,-LINF}}); }
    void upd(int node, Line v) {
        auto [l,r,s,e,curl]=tree[node];
        ll m=(s+e)>>1;
        Line lo=curl, hi=v;

        if (lo.get(s)>hi.get(s)) swap(lo,hi);
        if (lo.get(e)<=hi.get(e)) {
            tree[node].line=hi;
            return;
        }

        if (lo.get(m)<=hi.get(m)) {
            tree[node].line=hi;
            if (tree[node].r==-1) {
                tree[node].r=sz(tree);
                tree.pb({-1,-1,m+1,e,{0,-LINF,-LINF}});
            }
            upd(tree[node].r,lo);
        }
        else {
            tree[node].line=lo;
            if (tree[node].l==-1) {
                tree[node].l=sz(tree);
                tree.pb({-1,-1,s,m,{0,-LINF,-LINF}});
            }
            upd(tree[node].l,hi);
        }
    }
    pll qry(int node, ll x) {
        if (node==-1) return {-LINF,-LINF};
        ll s=tree[node].s, e=tree[node].e;
        ll m=(s+e)>>1;
        if (x<=m) return max(tree[node].line.get(x), qry(tree[node].l,x));
        else return max(tree[node].line.get(x), qry(tree[node].r,x));
    }
};

int N,M;
ll P[MAXN],D[2][MAXN];
int par[MAXM][MAXN];

int main() {
    fio();
    cin>>N>>M;
    for (int i=1; i<=N; i++) cin>>P[i], P[i]+=P[i-1];

    for (int j=1; j<=M; j++) {
        LiChao lic; lic.init(0,1e9+1);
        for (int i=1; i<=N; i++) {
            lic.upd(0,{P[i-1],-P[i-1]*P[i-1]+D[(j-1)&1][i-1],i-1});
            auto [mx,mxi]=lic.qry(0,P[i]);
            D[j&1][i]=mx;
            par[j][i]=mxi;
        }
    }

    cout<<D[M&1][N]<<endl;
    vector<int> idx;
    for (int x=N,y=M; y>=1; y--) {
        idx.pb(par[y][x]);
        x=par[y][x];
    }
    reverse(all(idx));
    for (int i:idx) cout<<i<<sp;
    return 0;
}

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