Submission #130143

#	Time	Username	Problem	Language	Result	Execution time	Memory
130143		mlyean00	Split the sequence (APIO14_sequence)	C++14	0 / 100	1587 ms	23984 KiB

sequence

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

struct line {
    int idx;
    ll a, b;

    line(int idx, ll a, ll b) : idx(idx), a(a), b(b) {}

    ll cost(ll x) const {
        return a * (x - a) + b;
    }

    ll intersect(line const& other) {
        ll lo = other.a;
        ll hi = 1LL << 48;

        while (lo < hi) {
            ll mid = (lo + hi) / 2;
            if (cost(mid) > other.cost(mid)) {
                lo = mid + 1;
            } else {
                hi = mid;
            }
        }

        return lo;
    }
};

int main() {
    int n, k;
    cin >> n >> k;

    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }

    vector<ll> pref(n + 1, 0);
    partial_sum(a.begin(), a.end(), pref.begin() + 1);

    ll dp[k + 1][n + 1];
    int parent[k + 1][n + 1];

    for (int i = 0; i <= k; ++i) {
        for (int j = 0; j <= n; ++j) {
            dp[i][j] = 0;
        }
    }

    for (int i = 1; i <= k; ++i) {
        deque<line> dq;
        for (int j = 0; j <= n; ++j) {
            while (dq.size() > 1 &&
                   dq[0].cost(pref[j]) <= dq[1].cost(pref[j])) {
                dq.pop_front();
            }
            if (!dq.empty()) {
                dp[i][j] = dq[0].cost(pref[j]);
                parent[i][j] = dq[0].idx;
            }
            line l(j, pref[j], dp[i - 1][j]);

            while (dq.size() > 1) {
                int sz = dq.size();
                ll t1 = dq[sz - 2].intersect(dq[sz - 1]);
                ll t2 = dq[sz - 1].intersect(l);

                if (t1 < t2) break;

                dq.pop_back();
            }
            dq.push_back(l);
        }
    }

    cout << dp[k][n] << endl;

    int u = n;
    for (int i = k; i > 0; --i) {
        u = parent[i][u];
        cout << u << ' ';
    }
    cout << endl;

    return 0;
}

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