이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <tuple>
#include <vector>
#include <deque>
using namespace std;
using ll = long long;
using line = tuple<ll, ll, int>;
const int N = 1e5;
const int K = 200;
int n, k;
ll a[N + 5], pfx_sum[N + 5];
pair<ll, int> dp[K + 5][N + 5];
line best[K + 5];
deque<line> q[K + 5];
ll ps(int i) {
return pfx_sum[i];
}
ll ss(int i) {
return pfx_sum[n] - pfx_sum[i - 1];
}
ll f(line l, ll x) {
auto [m, c, i] = l;
return m * x + c;
}
bool lastIsBad(line lz, line ly, line lx) {
auto [mz, cz, iz] = lz;
auto [my, cy, iy] = ly;
auto [mx, cx, ix] = lx;
return (cy - cx) * (mx - mz) <= (cz - cx) * (mx - my); // ((cy - cx) / (mx - my)) <= ((cz - cx) / (mx - mz))
}
pair<ll, vector<int>> retrieve() {
ll x = -1;
vector<int> v = {-1};
for (int i = 1; i <= n; i++) {
if (dp[k][i].first > x) {
x = dp[k][i].first;
v.back() = i;
}
}
for (int i = k; i >= 2; i--) {
v.push_back(dp[i][v.back()].second);
}
return {x, v};
}
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];
}
for (int i = 1; i <= n; i++) {
for (int s = k; s >= 1; s--) {
dp[s][i] = {-1, -1};
while (q[s - 1].size() > 1 and f(q[s - 1][0], ss(i + 1)) <= f(q[s - 1][1], ss(i + 1))) q[s - 1].pop_front();
if (s == 1) dp[s][i].first = ps(i) * ss(i + 1);
else if (not q[s - 1].empty()) {
dp[s][i].first = ps(i) * ss(i + 1) + f(q[s - 1][0], ss(i + 1));
dp[s][i].second = get<2>(q[s - 1][0]);
}
// cout << i << ' ' << s << ' ' << dp[s][i].first << ' ' << dp[s][i].second << endl;
if (s != 1 and dp[s][i].first == -1) continue;
line cur = make_tuple(-ps(i), dp[s][i].first, i);
while (q[s].size() > 1 and lastIsBad(q[s][q[s].size() - 1], q[s][q[s].size() - 2], cur)) q[s].pop_back();
q[s].push_back(cur);
}
}
pair<ll, vector<int>> ans = retrieve();
cout << ans.first << endl;
for (int i : ans.second) {
cout << i << ' ';
}
cout << endl;
return 0;
}
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