제출 #1088306

#제출 시각아이디문제언어결과실행 시간메모리
1088306rlx0090수열 (APIO14_sequence)C++14
22 / 100
2082 ms89232 KiB
#include <iostream> #include <string> #include <vector> #include <algorithm> #include <cmath> #include <limits> #include <queue> #include <string.h> #include <stack> #include <limits> using namespace std; int N, K; long long p[100005], dp[201][100005]; long long minf = numeric_limits<long long>::min(); struct seg { long long k = 0, b = minf; seg() {} seg(long long _k, long long _b) :k(_k), b(_b) {} long long f(long long x) { return k * x + b; } }; struct Node { Node * left = nullptr; Node * right = nullptr; seg s; ~Node() { if(left != nullptr) delete left; if(right != nullptr) delete right; } }; void add(Node* cur, seg ns, int nl, int nr) { if(cur->s.b == minf && ns.b == minf) return; int m = (nl + nr) / 2; bool lef = cur->s.f(nl) < ns.f(nl); bool mid = cur->s.f(m) < ns.f(m); if(mid) swap(cur->s, ns); if(nl == nr) return; else if(lef != mid) { if(cur->left == nullptr) cur->left = new Node(); add(cur->left, ns, nl, m); } else { if(cur->right == nullptr) cur->right = new Node(); add(cur->right, ns, m + 1, nr); } } long long query(Node* cur, int x, int nl, int nr) { if(cur == nullptr) return 0; if(nl == nr) return cur->s.f(x); int m = (nl + nr) / 2; if(x <= m) return max(cur->s.f(x), query(cur->left, x, nl, m)); else return max(cur->s.f(x), query(cur->right, x, m + 1, nr)); } int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cin >> N >> K; long long q; for(int i = 1; i <= N; ++i) { cin >> q; p[i] = p[i - 1] + q; } Node * tree; long long ans = 0; for(int k = 1; k <= K; ++k) { tree = new Node(); for(int i = N - k - 1; i >= 0; --i){ add(tree, seg(p[i + 1], - p[i + 1] * p[i + 1] + p[N] * p[i + 1] + dp[k - 1][i + 1]), 0, 1000 * N); dp[k][i] = query(tree, p[i], 0, 1000 * N) - p[N] * p[i]; } delete(tree); } ans = dp[K][0]; cout << ans << '\n'; int s = 0; for(int k = K; k >= 1; --k) { for(int i = s + 1; i <= N - k; ++i) { long long q = (p[N] - p[i]) * (p[i] - p[s]); if(ans - q == dp[k - 1][i] ) { cout << i << ' '; ans -= q; s = i; break; } } } return 0; }
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