제출 #847376

#제출 시각아이디문제언어결과실행 시간메모리
847376I_love_Hoang_Yen추월 (IOI23_overtaking)C++17
0 / 100
1 ms348 KiB
#include "overtaking.h" #include <bits/stdc++.h> using namespace std; using ll = long long; constexpr int MAX_SS = 1011; // Lazy Segment Tree, copied from AtCoder {{{ // Source: https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp // Doc: https://atcoder.github.io/ac-library/master/document_en/lazysegtree.html // // Notes: // - Index of elements from 0 // - Range queries are [l, r-1] // - composition(f, g) should return f(g()) // // Tested: // - https://oj.vnoi.info/problem/qmax2 // - https://oj.vnoi.info/problem/lites // - (range set, add, mult, sum) https://oj.vnoi.info/problem/segtree_itmix // - (range add (i-L)*A + B, sum) https://oj.vnoi.info/problem/segtree_itladder // - https://atcoder.jp/contests/practice2/tasks/practice2_l // - https://judge.yosupo.jp/problem/range_affine_range_sum int ceil_pow2(int n) { int x = 0; while ((1U << x) < (unsigned int)(n)) x++; return x; } template< class S, // node data type S (*op) (S, S), // combine 2 nodes S (*e) (), // identity element class F, // lazy propagation tag S (*mapping) (F, S), // apply tag F on a node F (*composition) (F, F), // combine 2 tags F (*id)() // identity tag > struct LazySegTree { LazySegTree() : LazySegTree(0) {} explicit LazySegTree(int n) : LazySegTree(vector<S>(n, e())) {} explicit LazySegTree(const vector<S>& v) : _n((int) v.size()) { log = ceil_pow2(_n); size = 1 << log; d = std::vector<S>(2 * size, e()); lz = std::vector<F>(size, id()); for (int i = 0; i < _n; i++) d[size + i] = v[i]; for (int i = size - 1; i >= 1; i--) { update(i); } } // 0 <= p < n void set(int p, S x) { assert(0 <= p && p < _n); p += size; for (int i = log; i >= 1; i--) push(p >> i); d[p] = x; for (int i = 1; i <= log; i++) update(p >> i); } // 0 <= p < n S get(int p) { assert(0 <= p && p < _n); p += size; for (int i = log; i >= 1; i--) push(p >> i); return d[p]; } // Get product in range [l, r-1] // 0 <= l <= r <= n // For empty segment (l == r) -> return e() S prod(int l, int r) { assert(0 <= l && l <= r && r <= _n); if (l == r) return e(); l += size; r += size; for (int i = log; i >= 1; i--) { if (((l >> i) << i) != l) push(l >> i); if (((r >> i) << i) != r) push((r - 1) >> i); } S sml = e(), smr = e(); while (l < r) { if (l & 1) sml = op(sml, d[l++]); if (r & 1) smr = op(d[--r], smr); l >>= 1; r >>= 1; } return op(sml, smr); } S all_prod() { return d[1]; } // 0 <= p < n void apply(int p, F f) { assert(0 <= p && p < _n); p += size; for (int i = log; i >= 1; i--) push(p >> i); d[p] = mapping(f, d[p]); for (int i = 1; i <= log; i++) update(p >> i); } // Apply f on all elements in range [l, r-1] // 0 <= l <= r <= n void apply(int l, int r, F f) { assert(0 <= l && l <= r && r <= _n); if (l == r) return; l += size; r += size; for (int i = log; i >= 1; i--) { if (((l >> i) << i) != l) push(l >> i); if (((r >> i) << i) != r) push((r - 1) >> i); } { int l2 = l, r2 = r; while (l < r) { if (l & 1) all_apply(l++, f); if (r & 1) all_apply(--r, f); l >>= 1; r >>= 1; } l = l2; r = r2; } for (int i = 1; i <= log; i++) { if (((l >> i) << i) != l) update(l >> i); if (((r >> i) << i) != r) update((r - 1) >> i); } } // Binary search on SegTree to find largest r: // f(op(a[l] .. a[r-1])) = true (assuming empty array is always true) // f(op(a[l] .. a[r])) = false (assuming op(..., a[n]), which is out of bound, is always false) template <bool (*g)(S)> int max_right(int l) { return max_right(l, [](S x) { return g(x); }); } template <class G> int max_right(int l, G g) { assert(0 <= l && l <= _n); assert(g(e())); if (l == _n) return _n; l += size; for (int i = log; i >= 1; i--) push(l >> i); S sm = e(); do { while (l % 2 == 0) l >>= 1; if (!g(op(sm, d[l]))) { while (l < size) { push(l); l = (2 * l); if (g(op(sm, d[l]))) { sm = op(sm, d[l]); l++; } } return l - size; } sm = op(sm, d[l]); l++; } while ((l & -l) != l); return _n; } // Binary search on SegTree to find smallest l: // f(op(a[l] .. a[r-1])) = true (assuming empty array is always true) // f(op(a[l-1] .. a[r-1])) = false (assuming op(a[-1], ..), which is out of bound, is always false) template <bool (*g)(S)> int min_left(int r) { return min_left(r, [](S x) { return g(x); }); } template <class G> int min_left(int r, G g) { assert(0 <= r && r <= _n); assert(g(e())); if (r == 0) return 0; r += size; for (int i = log; i >= 1; i--) push((r - 1) >> i); S sm = e(); do { r--; while (r > 1 && (r % 2)) r >>= 1; if (!g(op(d[r], sm))) { while (r < size) { push(r); r = (2 * r + 1); if (g(op(d[r], sm))) { sm = op(d[r], sm); r--; } } return r + 1 - size; } sm = op(d[r], sm); } while ((r & -r) != r); return 0; } private: int _n, size, log; vector<S> d; vector<F> lz; void update(int k) { d[k] = op(d[2*k], d[2*k+1]); } void all_apply(int k, F f) { d[k] = mapping(f, d[k]); if (k < size) lz[k] = composition(f, lz[k]); } void push(int k) { all_apply(2*k, lz[k]); all_apply(2*k+1, lz[k]); lz[k] = id(); } }; // }}} ll op(ll x, ll y) { return max(x, y); } ll e() { return 0ll; } ll mapping(ll f, ll s) { return max(s, f); } ll composition(ll f, ll g) { return max(f, g); } ll id() { return 0ll; } using STMax = LazySegTree<ll, op, e, ll, mapping, composition, id>; STMax st; // lines[i]: stores lines between i-th and (i+1)-th sorting stations // each line is represented by its 2 endpoints set<pair<ll, ll>> lines[MAX_SS]; int M; ll X; vector<int> S; vector<ll> xs; void init( int L, int nBus, vector<ll> T, vector<int> W, int _X, int _M, vector<int> _S) { // subtask 4 {{{ M = _M; X = _X; S = _S; // sort buses in decreasing order of W (so slowest buses are processed first) vector<pair<ll, ll>> buses; for (int i = 0; i < nBus; ++i) { if (W[i] <= X) continue; buses.emplace_back(W[i], T[i]); } std::sort(buses.begin(), buses.end()); std::reverse(buses.begin(), buses.end()); // init gaps between 2 sorting stations for (int i = 0; i < M-1; ++i) { // only M-1 gaps lines[i].clear(); lines[i].insert({0, 0}); } // for each bus, add its line to all gaps for (const auto& [w, t] : buses) { ll cur_time = t; for (int j = 0; j < M-1; ++j) { auto it = std::prev(lines[j].lower_bound({cur_time, 0})); ll exit_time = std::max(it->second, cur_time + w*(S[j+1] - S[j])); lines[j].insert({cur_time, exit_time}); cur_time = exit_time; } } // }}} for (int i = M-2; i >= 0; --i) { ll X_at_left = X * S[i]; ll X_at_right = X * S[i+1]; for (auto [l, r] : lines[i]) { if (!l && !r) continue; l -= X_at_left; r -= X_at_right; xs.push_back(l + 1); xs.push_back(r); } } std::sort(xs.begin(), xs.end()); xs.erase(unique(xs.begin(), xs.end()), xs.end()); st = STMax(xs); for (int i = M-2; i >= 0; --i) { ll X_at_left = X * S[i]; ll X_at_right = X * S[i+1]; for (auto [l, r] : lines[i]) { if (!l && !r) continue; l -= X_at_left; r -= X_at_right; st.apply( lower_bound(xs.begin(), xs.end(), l+1) - xs.begin(), lower_bound(xs.begin(), xs.end(), r) - xs.begin() + 1, r); } } } ll arrival_time(ll Y) { ll tx = X * S.back(); if (xs.empty() || Y <= xs[0] || Y > xs.back()) return tx + Y; return tx + st.get(lower_bound(xs.begin(), xs.end(), Y) - xs.begin() - 1); }
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