はじめに

AtCoder Problems の Recommendation を利用して、過去の問題を解いています。
AtCoder さん、AtCoder Problems さん、ありがとうございます。

今回のお題

AtCoder Beginner Contest D - Summer Vacation
Difficulty:1313

今回のテーマ、優先度付きキュー

ac-library-rb は、AtCoder Library (ACL)のRuby版です。

その ac-library-rb がなんとGem化されています。
ac-library-rb (rubygems)
AtCoderにインストールされる日が待ち遠しいですね。

優先度付きキュー

ruby.rb

# Priority Queue
# Reference: https://github.com/python/cpython/blob/master/Lib/heapq.py
class PriorityQueue
  # By default, the priority queue returns the maximum element first.
  # If a block is given, the priority between the elements is determined with it.
  # For example, the following block is given, the priority queue returns the minimum element first.
  # `PriorityQueue.new { |x, y| x < y }`
  #
  # A heap is an array for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for all k, counting elements from 0.
  def initialize(array = [], &comp)
    @heap = array
    @comp = comp || proc { |x, y| x > y }
    heapify
  end

  attr_reader :heap

  # Push new element to the heap.
  def push(item)
    shift_down(0, @heap.push(item).size - 1)
  end

  alias << push
  alias append push

  # Pop the element with the highest priority.
  def pop
    latest = @heap.pop
    return latest if empty?

    ret_item = heap[0]
    heap[0] = latest
    shift_up(0)
    ret_item
  end

  # Get the element with the highest priority.
  def get
    @heap[0]
  end

  alias top get

  # Returns true if the heap is empty.
  def empty?
    @heap.empty?
  end

  private

  def heapify
    (@heap.size / 2 - 1).downto(0) { |i| shift_up(i) }
  end

  def shift_up(pos)
    end_pos = @heap.size
    start_pos = pos
    new_item = @heap[pos]
    left_child_pos = 2 * pos + 1

    while left_child_pos < end_pos
      right_child_pos = left_child_pos + 1
      if right_child_pos < end_pos && @comp.call(@heap[right_child_pos], @heap[left_child_pos])
        left_child_pos = right_child_pos
      end
      # Move the higher priority child up.
      @heap[pos] = @heap[left_child_pos]
      pos = left_child_pos
      left_child_pos = 2 * pos + 1
    end
    @heap[pos] = new_item
    shift_down(start_pos, pos)
  end

  def shift_down(star_pos, pos)
    new_item = @heap[pos]
    while pos > star_pos
      parent_pos = (pos - 1) >> 1
      parent = @heap[parent_pos]
      break if @comp.call(parent, new_item)

      @heap[pos] = parent
      pos = parent_pos
    end
    @heap[pos] = new_item
  end
end

n, m = gets.split.map(&:to_i)
ab = Array.new(100000 + 1){ [] }
n.times do
  a, b = gets.split.map(&:to_i)
  ab[a] << b
end
ans = 0
pq = PriorityQueue.new
1.upto(m) do |i|
  next if ab[i].nil?
  ab[i].each do |e|
    pq << e
  end
  ans += pq.pop if !pq.empty?
end
puts ans

今はインストールされていないのでclass部分を追記する必要があります。

upto.rb

1.upto(m) do |i|

最終日に近い日から貪欲に選択します。

まとめ

優先度付きキュー(Priority Queue) を解いた
ACL に詳しくなった
Ruby に詳しくなった