ruby.rb

class UnionFind
  def initialize(n)
    @parents = Array.new(n, -1)
  end
  def find(x)
    @parents[x] < 0 ? x : @parents[x] = find(@parents[x])
  end
  def parents
    @parents
  end
  def union(x, y)
    x = find(x)
    y = find(y)
    return if x == y
    if @parents[x] > @parents[y]
      x, y = y, x
    end
    @parents[x] += @parents[y]
    @parents[y] = x
  end
end

n, m = gets.split.map(&:to_i)
u = UnionFind.new(n)
m.times do
  a, b = gets.split.map(&:to_i)
  u.union(a - 1, b - 1)
end
puts u.parents.select{ _1 < 0 }.size - 1

sub.rb

puts u.parents.select{ _1 < 0 }.size - 1 #this time

puts -u.parents.min #Last time

Just change the last line of Last time --Qiita to ʻAC`.

sub.rb

@parents = [-3, 0, -2, 2, 0]

Explaining in the previous input example 1, in ʻu.parents, ʻu.parents [0] and ʻu.parents [2]` can be divided into two representative groups. It's OK if you connect it with one road.

Ruby version of AtCoder Library (ACL)

There is also a move to create a Ruby version of ACL on here --github. It's attention required.

Summary

• Solved ACLBC C
• Become familiar with Ruby

Referenced site

• Solving with Ruby AtCoder ABC177 D UnionFind --Qiita *