제출 #1261670

#제출 시각아이디문제언어결과실행 시간메모리
1261670biank팀들 (IOI15_teams)C++20
100 / 100
914 ms183592 KiB
#include "teams.h" #include <bits/stdc++.h> #define forsn(i, s, n) for (int i = int(s); i < int(n); i++) #define forn(i, n) forsn(i, 0, n) #define dforsn(i, s, n) for (int i = int(n) - 1; i >= int(s); i--) #define dforn(i, n) dforsn(i, 0, n) #define sz(x) int(x.size()) #define all(x) begin(x), end(x) #define pb push_back #define eb emplace_back #define fst first #define snd second using namespace std; using vi = vector<int>; using ii = pair<int, int>; const int SZ = 1 << 19; vi st(1, 0), l(1, 0), r(1, 0); int newNode(int u) { st.pb(st[u]), l.pb(l[u]), r.pb(r[u]); return sz(st) - 1; } int update(int node, int pos, int val, int lo = 0, int hi = SZ) { node = newNode(node); if (hi - lo == 1) { st[node] += val; return node; } int mid = (lo + hi) / 2; if (pos < mid) { l[node] = update(l[node], pos, val, lo, mid); } else { r[node] = update(r[node], pos, val, mid, hi); } st[node] = st[l[node]] + st[r[node]]; return node; } int query(int s, int e, int node1, int node2, int lo, int hi) { if (e <= lo || hi <= s) return 0; if (s <= lo && hi <= e) return st[node2] - st[node1]; int mid = (lo + hi) / 2; return query(s, e, l[node1], l[node2], lo, mid) + query(s, e, r[node1], r[node2], mid, hi); } int find(int need, int node1, int node2, int lo, int hi) { if (hi - lo == 1) return lo; int mid = (lo + hi) / 2; int leftSum = st[l[node2]] - st[l[node1]]; if (leftSum >= need) { return find(need, l[node1], l[node2], lo, mid); } return find(need - leftSum, r[node1], r[node2], mid, hi); } int n; int roots[SZ]; int query(int l1, int r1, int l2, int r2) { return query(l1, r1, roots[l2 - 1], roots[r2 - 1], 0, SZ); } int find(int k, int l, int r) { return find(k, roots[l - 1], roots[r - 1], 0, SZ); } void init(int N, int A[], int B[]) { n = N; st.reserve(1e7), l.reserve(1e7), r.reserve(1e7); vector<vi> byA(n + 1); forn(i, n) byA[A[i]].pb(B[i]); roots[0] = 0; forsn(i, 1, n + 1) { roots[i] = roots[i - 1]; for (int b : byA[i]) roots[i] = update(roots[i], b, 1); } } int can(int M, int K[]) { vi a, b, cnt; sort(K, K + M); a.pb(0), b.pb(n), cnt.pb(0); forn(i, M) { while (!b.empty() && b.back() < K[i]) { a.pop_back(), b.pop_back(), cnt.pop_back(); } int prevB = K[i], need = K[i]; while (!cnt.empty()) { int total = query(prevB, b.back() + 1, a.back() + 1, K[i] + 1) + cnt.back(); if (total < need) { need -= total, prevB = b.back() + 1; a.pop_back(), b.pop_back(), cnt.pop_back(); } else { break; } } if (cnt.empty()) return 0; int prev = query(0, prevB, a.back() + 1, K[i] + 1); int minB = find(prev + need, a.back() + 1, K[i] + 1); minB = min(minB, b.back()); int total = query(prevB, minB + 1, a.back() + 1, K[i] + 1); if (minB == b.back()) total += cnt.back(), cnt.back() = 0; assert(total >= need); a.pb(K[i]), b.pb(minB), cnt.pb(total - need); } return 1; }
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