제출 #725424

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
725424		colossal_pepe	수열 (APIO14_sequence)	C++17	33 / 100	3 ms	724 KiB

sequence

#include <iostream>
#include <deque>
#include <algorithm>
#include <cstring>
using namespace std;

//    l[n] is bad if l[n - 1] intersects l[n + 1] before l[n] does
// => l[0] is bad if l[1] intersects l[-1] before l[0] does

using ll = long long;

const int N = 500;
const int K = 200;

int n, k, par[K + 5][N + 5];
ll a[N + 5], pfx_sum[N + 5], dp[N + 5];
deque<int> l;

ll ps(int i) {
    return pfx_sum[i];
}

ll ss(int i) {
    return pfx_sum[n] - pfx_sum[i - 1];
}

ll f(int li, ll x) {
    return -ps(li) * x + dp[li];
}

bool firstIsBad(int l1, int l2, int l0) {
    ll m1 = -ps(l1), c1 = dp[l1];
    ll m2 = -ps(l2), c2 = dp[l2];
    ll m0 = -ps(l0), c0 = dp[l0];
    return (c2 - c0) * (m0 - m1) > (c1 - c0) * (m0 - m2); // ((c2 - c0) / (m0 - m2)) > ((c1 - c0) / (m0 - m1))
}

void dfs(int i, int s) {
    cout << i << ' ';
    if (s > 1) dfs(par[s][i], s - 1);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n >> k;
    a[0] = pfx_sum[0] = 0;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        pfx_sum[i] = a[i] + pfx_sum[i - 1];
    }
    memset(par, -1, sizeof par);
    for (int s = 1; s <= k; s++) {
        deque<int> l_new;
        for (int i = n; i >= 1; i--) {
            dp[i] = -1;
            while (l.size() > 1 and (i <= l[0] or f(l[0], ss(i + 1)) <= f(l[1], ss(i + 1)))) l.pop_front();
            if (s == 1) dp[i] = ps(i) * ss(i + 1);
            else if (not l.empty() and l[0] < i) {
                dp[i] = ps(i) * ss(i + 1) + f(l[0], ss(i + 1));
                par[s][i] = l[0];
            }
            if (dp[i] == -1) continue;
            while (l_new.size() > 1 and firstIsBad(l_new[0], l_new[1], i)) l_new.pop_front();
            l_new.push_front(i);
        }
        reverse(l_new.begin(), l_new.end());
        l = l_new;
    }
    ll ans = 1;
    for (int i = 1; i <= n; i++) {
        if (dp[ans] < dp[i]) ans = i;
    }
    cout << dp[ans] << '\n';
    dfs(ans, k);
    cout << '\n';
    return 0;
}

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