lmori's Library

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:heavy_check_mark: Longest Increasing Subsequence
(dp/LongestIncreasingSubsequence.hpp)

概要

todo

計算量

todo

Depends on

Verified with

Code

#pragma once
#include "../data-structure/segment-tree/SegmentTree.hpp"
template <typename T>
int lis(vector<T> a) {
  using S = int;
  auto op = [](S x, S y) { return max(x, y); };
  auto e = []() { return 0; };
  int n = a.size();
  int ret = 0;
  vector<T> b(n);
  b = a;
  sort(b.begin(), b.end());
  b.erase(unique(b.begin(), b.end()), b.end());
  for (int i = 0; i < n; i++) a[i] = distance(b.begin(), lower_bound(b.begin(), b.end(), a[i]));
  int siz = b.size();
  segtree<S, op, e> dp(siz);
  for (int i = 0; i < n; i++) {
    int cur = max(1, dp.prod(0, a[i]) + 1);
    dp.set(a[i], cur);
    ret = max(ret, cur);
  }
  return ret;
}
#line 2 "data-structure/segment-tree/SegmentTree.hpp"
template <class S, auto op, auto e>
struct segtree {
 private:
  unsigned int seg_bit_ceil(unsigned int n) {
    unsigned int x = 1;
    while (x < (unsigned int)(n)) x *= 2;
    return x;
  }

 public:
  static_assert(std::is_convertible_v<decltype(op), std::function<S(S, S)>>,
                "op must work as S(S, S)");
  static_assert(std::is_convertible_v<decltype(e), std::function<S()>>,
                "e must work as S()");
  segtree() : segtree(0) {}
  explicit segtree(int n) : segtree(std::vector<S>(n, e())) {}
  explicit segtree(const std::vector<S>& v) : _n(int(v.size())) {
    size = (int)seg_bit_ceil((unsigned int)(_n));
    log = __builtin_ctz((unsigned int)size);
    d = std::vector<S>(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);
    }
  }

  void set(int p, S x) {
    assert(0 <= p && p < _n);
    p += size;
    d[p] = x;
    for (int i = 1; i <= log; i++) update(p >> i);
  }

  S get(int p) const {
    assert(0 <= p && p < _n);
    return d[p + size];
  }

  S prod(int l, int r) const {
    assert(0 <= l && l <= r && r <= _n);
    S 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);
  }

  S all_prod() const { return d[1]; }

  template <bool (*f)(S)>
  int max_right(int l) const {
    return max_right(l, [](S 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;
    S 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;
  }

  template <bool (*f)(S)>
  int min_left(int r) const {
    return min_left(r, [](S 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;
    S 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;
  std::vector<S> d;

  void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); }
};
#line 3 "dp/LongestIncreasingSubsequence.hpp"
template <typename T>
int lis(vector<T> a) {
  using S = int;
  auto op = [](S x, S y) { return max(x, y); };
  auto e = []() { return 0; };
  int n = a.size();
  int ret = 0;
  vector<T> b(n);
  b = a;
  sort(b.begin(), b.end());
  b.erase(unique(b.begin(), b.end()), b.end());
  for (int i = 0; i < n; i++) a[i] = distance(b.begin(), lower_bound(b.begin(), b.end(), a[i]));
  int siz = b.size();
  segtree<S, op, e> dp(siz);
  for (int i = 0; i < n; i++) {
    int cur = max(1, dp.prod(0, a[i]) + 1);
    dp.set(a[i], cur);
    ret = max(ret, cur);
  }
  return ret;
}
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