제출 #286454

#제출 시각아이디문제언어결과실행 시간메모리
286454dolphingarlic모임들 (IOI18_meetings)C++14
36 / 100
1060 ms11000 KiB
#include "meetings.h" #include <bits/stdc++.h> typedef long long ll; using namespace std; struct Node { int l, r, a, b, tot; Node operator+(Node B) { Node ret = {l, B.r, (tot == r - l + 1 ? tot + B.a : a), (B.tot == B.r - B.l + 1 ? b + B.tot : B.b), max({tot, B.tot, b + B.a})}; return ret; } } segtree[400001]; int n, h[100001]; ll dp[5000]; void build(int node = 1, int l = 0, int r = n - 1) { if (l == r) segtree[node] = {l, r, h[l] == 1, h[l] == 1, h[l] == 1}; else { int mid = (l + r) / 2; build(node * 2, l, mid); build(node * 2 + 1, mid + 1, r); segtree[node] = segtree[node * 2] + segtree[node * 2 + 1]; } } Node query(int a, int b, int node = 1, int l = 0, int r = n - 1) { if (a <= l && b >= r) return segtree[node]; int mid = (l + r) / 2; if (mid >= a) { if (mid + 1 <= b) return query(a, b, node * 2, l, mid) + query(a, b, node * 2 + 1, mid + 1, r); return query(a, b, node * 2, l, mid); } return query(a, b, node * 2 + 1, mid + 1, r); } vector<ll> minimum_costs(vector<int> H, vector<int> L, vector<int> R) { n = H.size(); for (int i = 0; i < n; i++) h[i] = H[i]; int Q = int(L.size()); vector<long long> C(Q); if (Q <= 5000) { for (int i = 0; i < Q; i++) { memset(dp, 0, sizeof dp); stack<pair<int, ll>> stck; ll val = 0; for (int j = L[i]; j <= R[i]; j++) { while (stck.size() && H[stck.top().first] <= H[j]) { val -= H[stck.top().first] * stck.top().second; stck.pop(); } ll rng = (stck.size() ? j - stck.top().first : j - L[i] + 1); val += rng * H[j]; dp[j] += val; stck.push({j, rng}); // cout << dp[j] << ' '; } // cout << "| "; while (stck.size()) stck.pop(); val = 0; for (int j = R[i]; j >= L[i]; j--) { while (stck.size() && H[stck.top().first] <= H[j]) { val -= H[stck.top().first] * stck.top().second; stck.pop(); } ll rng = (stck.size() ? stck.top().first - j : R[i] - j + 1); val += rng * H[j]; dp[j] += val - H[j]; stck.push({j, rng}); // cout << dp[j] << ' '; } // cout << '\n'; C[i] = *min_element(dp + L[i], dp + R[i] + 1); } } else { build(); for (int i = 0; i < Q; i++) { C[i] = 2 * (R[i] - L[i] + 1) - query(L[i], R[i]).tot; } } return C; }
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