Submission #580048

#TimeUsernameProblemLanguageResultExecution timeMemory
580048leeh18Split the sequence (APIO14_sequence)C++17
100 / 100
1070 ms95708 KiB
// #pragma GCC optimize("O3") #include <bits/stdc++.h> using namespace std; using ll = long long; #define sz(x) (int)size(x) #define all(x) begin(x), end(x) #define rall(x) rbegin(x), rend(x) #define rep(i, a, b) for(int i = a; i < (b); ++i) typedef pair<int, int> pii; typedef vector<int> vi; // https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0200r0.html template<class Fun> class y_combinator_result { Fun fun_; public: template<class T> explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {} template<class ...Args> decltype(auto) operator()(Args &&...args) { return fun_(std::ref(*this), std::forward<Args>(args)...); } }; template<class Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); } mt19937 rng((unsigned)chrono::steady_clock::now().time_since_epoch().count()); struct line_container { struct line { ll a, b, idx; line(ll a_, ll b_, int idx_) : a(a_), b(b_), idx(idx_) {} line() : a(0LL), b(0LL), idx(0) {} ll operator()(ll x) const { return a * x + b; } }; ll floor_div(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); } ll intersect(const line& l1, const line& l2) { return floor_div(l2.b - l1.b, l1.a - l2.a); } deque<line> dq; void add(ll a, ll b, int idx) { line cur(a, b, idx); if (!dq.empty() && dq.back().a == a) { if (dq.back().b > b) { return; } else { dq.pop_back(); } } while (sz(dq) > 1 && intersect(cur, dq[sz(dq)-1]) <= intersect(dq[sz(dq)-1], dq[sz(dq)-2])) { dq.pop_back(); } dq.push_back(cur); } pair<ll, int> query(ll x) { assert((!dq.empty())); while (sz(dq) > 1 && dq[0](x) < dq[1](x)) { dq.pop_front(); } return {dq[0](x), dq[0].idx}; } }; void solve() { int n, k; cin >> n >> k; vector<ll> p(n+1); for (int i = 1; i <= n; i++) { cin >> p[i]; p[i] += p[i-1]; } vector<line_container> g(k); vector<vector<int>> parent(n+1, vector<int>(k+1)); vector<ll> dp(k+1); for (int i = 1; i <= n; i++) { dp[0] = 0; for (int j = 1; j < i && j <= k; j++) { tie(dp[j], parent[i][j]) = g[j-1].query(p[i]); } for (int j = 0; j < i && j < k; j++) { g[j].add(p[i], -p[i]*p[i] + dp[j], i); } } cout << dp[k] << '\n'; vector<int> result; int idx = n; for (int i = k; i >= 1; i--) { result.push_back(parent[idx][i]); idx = parent[idx][i]; } for (int i = k-1; i >= 0; i--) { cout << result[i] << ' '; } } signed main() { cin.tie(0)->sync_with_stdio(0); cin.exceptions(cin.failbit); int t = 1; while (t--) { solve(); } return 0; }
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