Submission #288326

#TimeUsernameProblemLanguageResultExecution timeMemory
288326andrewMeetings (IOI18_meetings)C++17
19 / 100
560 ms196476 KiB
#include <bits/stdc++.h> #include "meetings.h" #define fi first #define se second #define p_b push_back #define m_p make_pair #define sz(x) (int)x.size() #define pii pair <int, int> #define all(x) x.begin(),x.end() using namespace std; typedef long long ll; const ll inf = 1e18; struct node{ int pref, suf, size, ans; }; vector <node> T; node clr; node merge(node a, node b){ if(a.size == 0)return b; if(b.size == 0)return a; node res; res.size = a.size + b.size; res.pref = a.pref; if(a.pref == a.size)res.pref = a.size + b.pref; res.suf = b.suf; if(b.suf == b.size)res.suf = b.size + a.suf; res.ans = max(a.ans, max(b.ans, a.suf + b.pref)); return res; } void build(int v, int tl, int tr, vector <int> &H){ if(tl == tr){ T[v].pref = T[v].suf = T[v].ans = (H[tl] == 2); T[v].size = 1; }else{ int tm = (tl + tr) >> 1; build(v << 1, tl, tm, H); build(v << 1 | 1, tm + 1, tr, H); T[v] = merge(T[v << 1], T[v << 1 | 1]); } } node get(int v, int tl, int tr, int l, int r){ if(l > r)return clr; if(tl == l && tr == r)return T[v]; int tm = (tl + tr) >> 1; return merge(get(v << 1, tl, tm, l, min(r, tm)), get(v << 1 | 1, tm + 1, tr, max(l, tm + 1), r)); } vector<ll> minimum_costs(vector<int> H, vector<int> l, vector<int> r) { int n = sz(H), q = sz(l); if(n <= 5000 && q <= 5000){ vector <ll> ans(q); vector < vector <ll> > pre_calc(n, vector <ll> (n, 0)); for(int x = 0; x < n; x++){ int mx = H[x]; ll sc = 0; for(int j = x; j < n; j++){ mx = max(mx, H[j]); sc += mx; pre_calc[x][j] = sc; } mx = H[x]; sc = 0; for(int j = x; j >= 0; j--){ mx = max(mx, H[j]); sc += mx; pre_calc[x][j] = sc; } } for(int i = 0; i < q; i++){ ans[i] = pre_calc[l[i]][r[i]]; for(int x = l[i] + 1; x <= r[i]; x++){ ans[i] = min(ans[i], pre_calc[x - 1][l[i]] + pre_calc[x][r[i]]); } } return ans; }else{ vector <ll> ans(q); T.resize(4 * n + 3); build(1, 0, n - 1, H); for(int i = 0; i < q; i++){ ans[i] = 2 * (r[i] - l[i] + 1) - get(1, 0, n - 1, l[i], r[i]).ans; } return ans; } }
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