Submission #682960

#	Time	Username	Problem	Language	Result	Execution time	Memory
682960		kusssso	Split the sequence (APIO14_sequence)	C++17	71 / 100	146 ms	131072 KiB

sequence

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const int N = 1e5 + 5;

struct line {
    ll A, B;
    bool operator < (const line &oth) const {
        return make_pair(A, B) < make_pair(oth.A, oth.B);
    }
    ld ins (line oth) {
        return 1.0 * (oth.B - B) / (A - oth.A);
    }
    ll yy (ll x) {
        return A * x + B;
    }
};

struct convex_hull { // finding maximum
    vector<line> hull;
    bool ok (line l1, line l2, line l3) {
        return l1.ins(l3) < l1.ins(l2);
    }
    void add (line d) {
        while (!hull.empty() && hull.back().A == d.A) hull.pop_back(); // hull.back().B < d.B
        while (hull.size() >= 2 && ok(hull[hull.size() - 2], hull[hull.size() - 1], d)) hull.pop_back();
        hull.push_back(d);
    }
    ll query (ll x) {
        int lo = 0, hi = (int) hull.size() - 2, p = 0;
        while (lo <= hi) {
           int mid = (lo + hi) / 2;
           if (hull[mid].ins(hull[mid + 1]) <= x) {
               p = mid + 1;
               lo = mid + 1;
           }
           else {
               hi = mid - 1;
           }
        }
        return hull[p].yy(x);
    }
} H;

int n, k, a[N];
ll f[N], dp[N][225];

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        f[i] = f[i - 1] + a[i];
    }
    for (int i = 1; i <= n; i++) {
        dp[i][1] = f[i] * (f[n] - f[i]);
    }
    k++;
    for (int t = 2; t <= k; t++) {
        convex_hull H;
        for (int i = 1; i <= n; i++) {
            H.add({f[i - 1], dp[i - 1][t - 1] - f[i - 1] * f[n]});
            dp[i][t] = max(dp[i][t], H.query(f[i]) + f[i] * (f[n] - f[i]));
            // for (int j = 1; j <= i; j++) {
                // // dp[i][t] = max(dp[i][t], dp[j - 1][t - 1] + (f[i] - f[j - 1]) * (f[n] - f[i])); 
                // dp[i][t] = max(dp[i][t], f[j - 1] * f[i] + (dp[j - 1][t - 1] - f[j - 1] * f[n]) + f[i] * (f[n] - f[i]));
                // //                           A       x                             B
            // }
        }
    }
    cout << dp[n][k] << '\n';
    vector<int> res;
    int i = n;
    for (int t = k; t >= 2; t--) {
        for (int j = i; j >= 1; j--) {
            if (dp[i][t] == dp[j - 1][t - 1] + (f[i] - f[j - 1]) * (f[n] - f[i])) {
                i = j - 1;
                res.push_back(i);
                break;
            }
        }
    }
    reverse(res.begin(), res.end());
    for (auto &e : res) cout << e << ' ';
    
    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