Submission #1269463

#TimeUsernameProblemLanguageResultExecution timeMemory
1269463rayan_bd수열 (APIO14_sequence)C++20
0 / 100
0 ms320 KiB
#include <bits/stdc++.h>
using namespace std;

const int mxN = 1e5 + 2;
const int mxK = 202;

#define ll long long

const ll inf = 1e18;

pair<ll, int> dp[mxN][mxK];
ll pref[mxN];

struct Line {
    mutable ll k, m, p, idx;
    bool operator<(const Line& o) const { return k < o.k; }
    bool operator<(int x) const { return p < x; }
};

struct LineContainer : multiset<Line, less<>> {
    static const ll inf = LLONG_MAX;
    ll div(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
    bool isect(iterator x, iterator y) {
        if (y == end()) { x->p = inf; return false; }
        if (x->k == y->k) x->p = x->m > y->m ? inf : -inf;
        else x->p = div(y->m - x->m, x->k - y->k);
        return x->p >= y->p;
    }
    void add(ll k, ll m, int idx) {
        auto z = insert({k, m, 0, idx}), y = z++, x = y;
        while (isect(y, z)) z = erase(z);
        if (x != begin() && isect(--x, y)) isect(x, y = erase(y));
        while ((y = x) != begin() && (--x)->p >= y->p)
            isect(x, erase(y));
    }
    pair<ll, int> query(ll x) {
        assert(!empty());
        auto l = *lower_bound(x);
        return {l.k * x + l.m, l.idx};
    }
} cht;


signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(nullptr);

    int n, K;
    cin >> n >> K;

    for(int i = 1; i <= n; ++i){
        cin >> pref[i];
        pref[i] += pref[i - 1];
    }

    for(int i = 0; i <= n + 1; ++i){
        for(int j = 0; j <= K; ++j){
            dp[i][j] = {inf, 0};
        }
    }

    for (int i = 1; i <= n + 1; i++) dp[i][0] = {0,  i};

    for (int k = 1; k <= K; k++) {
        cht.clear();
        for (int j = n; j >= 1; j--) {
            cht.add(pref[j], (long long) dp[j+1][k - 1].first - pref[j] * pref[j], j);
            auto [val, idx] = cht.query((long long) pref[n] + pref[j-1]);
            dp[j][k] = {(long long) -pref[j-1] * pref[n] + val, idx};
        }
    }

    cout << dp[1][K].first << "\n";

    int i = 1;
    while (K > 0 && i <= n) {
        int j = dp[i][K].second;
        if (j == -1) break;
        cout << j << " ";
        i = j + 1;
        K--;
    }

    cout << "\n";

    return 0;
}
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