제출 #753015

#제출 시각아이디문제언어결과실행 시간메모리
753015I_love_Hoang_Yen서열 (APIO23_sequence)C++17
40 / 100
2081 ms42144 KiB
#include "sequence.h" #include <bits/stdc++.h> #define SZ(s) ((int) ((s).size())) using namespace std; // SegTree, copied from AtCoder library {{{ // AtCoder doc: https://atcoder.github.io/ac-library/master/document_en/segtree.html // // Notes: // - Index of elements from 0 -> n-1 // - Range queries are [l, r-1] // // Tested: // - (binary search) https://atcoder.jp/contests/practice2/tasks/practice2_j // - https://oj.vnoi.info/problem/gss // - https://oj.vnoi.info/problem/nklineup // - (max_right & min_left for delete position queries) https://oj.vnoi.info/problem/segtree_itstr // - https://judge.yosupo.jp/problem/point_add_range_sum // - https://judge.yosupo.jp/problem/point_set_range_composite int ceil_pow2(int n) { int x = 0; while ((1U << x) < (unsigned int)(n)) x++; return x; } template< class T, // data type for nodes T (*op) (T, T), // operator to combine 2 nodes T (*e)() // identity element > struct SegTree { SegTree() : SegTree(0) {} explicit SegTree(int n) : SegTree(vector<T> (n, e())) {} explicit SegTree(const vector<T>& v) : _n((int) v.size()) { log = ceil_pow2(_n); size = 1<<log; d = vector<T> (2*size, e()); 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, T x) { assert(0 <= p && p < _n); p += size; d[p] = x; for (int i = 1; i <= log; i++) update(p >> i); } // 0 <= p < n T get(int p) const { assert(0 <= p && p < _n); return d[p + size]; } // Get product in range [l, r-1] // 0 <= l <= r <= n // For empty segment (l == r) -> return e() T prod(int l, int r) const { assert(0 <= l && l <= r && r <= _n); T sml = e(), smr = e(); l += size; r += size; 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); } T all_prod() const { return d[1]; } // 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 (*f)(T)> int max_right(int l) const { return max_right(l, [](T x) { return f(x); }); } template <class F> int max_right(int l, F f) const { assert(0 <= l && l <= _n); assert(f(e())); if (l == _n) return _n; l += size; T sm = e(); do { while (l % 2 == 0) l >>= 1; if (!f(op(sm, d[l]))) { while (l < size) { l = (2 * l); if (f(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 (*f)(T)> int min_left(int r) const { return min_left(r, [](T x) { return f(x); }); } template <class F> int min_left(int r, F f) const { assert(0 <= r && r <= _n); assert(f(e())); if (r == 0) return 0; r += size; T sm = e(); do { r--; while (r > 1 && (r % 2)) r >>= 1; if (!f(op(d[r], sm))) { while (r < size) { r = (2 * r + 1); if (f(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<T> d; void update(int k) { d[k] = op(d[2*k], d[2*k+1]); } }; // }}} // SegTree examples {{{ // Examples: Commonly used SegTree ops: max / min / sum struct MaxSegTreeOp { static int op(int x, int y) { return max(x, y); } static int e() { return INT_MIN; } }; struct MinSegTreeOp { static int op(int x, int y) { return min(x, y); } static int e() { return INT_MAX; } }; struct SumSegTreeOp { static long long op(long long x, long long y) { return x + y; } static long long e() { return 0; } }; // using STMax = SegTree<int, MaxSegTreeOp::op, MaxSegTreeOp::e>; // using STMin = SegTree<int, MinSegTreeOp::op, MinSegTreeOp::e>; // using STSum = SegTree<int, SumSegTreeOp::op, SumSegTreeOp::e>; // }}} bool can(int n, int eq, const vector<int>& a, const vector<vector<int>>& ids) { int ln = *max_element(a.begin(), a.end()); for (int median = 0; median <= ln; median++) { if (SZ(ids[median]) < eq) continue; vector<int> f(n+1, 0); for (int i = 1; i <= n; ++i) { f[i] = max(-1, min(a[i] - median, 1)); } std::partial_sum(f.begin(), f.end(), f.begin()); SegTree<int, MaxSegTreeOp::op, MaxSegTreeOp::e> st_max(f); SegTree<int, MinSegTreeOp::op, MinSegTreeOp::e> st_min(f); for (int ix = 0, iy = eq-1; iy < SZ(ids[median]); ++ix, ++iy) { int x = ids[median][ix]; int y = ids[median][iy]; // find [l, r]: // - l <= x < y <= r // - less + eq >= greater // - greater + eq >= less // - eq >= greater - less >= -eq // - eq >= (greater(r) - less(r)) - (greater(l-1) - less(l-1)) >= -eq int max_val = st_max.prod(y, n+1) - st_min.prod(0, x); int min_val = st_min.prod(y, n+1) - st_max.prod(0, x); // [-eq, eq] and [min_val, max_val] intersects if (min_val <= eq && max_val >= -eq) return true; } } return false; } int sequence(int n, std::vector<int> a) { // ids from 1 a.insert(a.begin(), 0); vector<vector<int>> ids(n + 1); for (int i = 1; i <= n; ++i) { ids[a[i]].push_back(i); } int left = 1, right = n, res = 1; while (left <= right) { int mid = (left + right) / 2; if (can(n, mid, a, ids)) { res = mid; left = mid + 1; } else { right = mid - 1; } } return res; }
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