Problem:
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Breadth First Search

Write a program which reads an directed graph $G = (V, E)$, and finds the shortest distance from vertex $1$ to each vertex (the number of edges in the shortest path). Vertices are identified by IDs $1, 2, ... n$.

Input

In the first line, an integer $n$ denoting the number of vertices, is given. In the next $n$ lines, adjacent lists of vertex $u$ are given in the following format:

$u$ $k$ $v_1$ $v_2$ ... $v_k$

$u$ is ID of the vertex and $k$ denotes its degree.$v_i$ are IDs of vertices adjacent to $u$.

Constraints

$1 \leq n \leq 100$

Output

For each vertex $u$, print $id$ and $d$ in a line. $id$ is ID of vertex $u$ and $d$ is the distance from vertex $1$ to vertex $u$. If there are no path from vertex $1$ to vertex $u$, print -1 as the shortest distance. Print in order of IDs.

Sample Input 1

4
1 2 2 4
2 1 4
3 0
4 1 3

Sample Output 1

1 0
2 1
3 2
4 1

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Ruby (judge verdict: Accepted)

class Vertex
  @@index = 0
  def initialize(adjacent)
    @index = @@index += 1
    @adjacent = adjacent
    @distance = -1
  end

  def to_s
    "#{index} #{distance}"
  end

  attr_reader :index
  attr_accessor :adjacent, :distance
end

def breadth_first_search(vertex)
  vertex.distance = 0
  queue = [vertex]
  until queue.empty?
    current = queue.shift
      current.adjacent.each do |x|
        if x.distance == -1
          x.distance = current.distance + 1
          queue << current.adjacent
          queue.flatten!
        end
      end
  end
end

n = gets.to_i
array = Array.new(n).map { Vertex.new([]) }
n.times do
  adjacent = []
  input = gets.split(' ').map(&:to_i)
  unless input[1].zero?
    input[2..-1].each do |index|
      adjacent << array.find { |vertex| vertex.index == index }
    end
    array[input[0] - 1].adjacent = adjacent
  end
end

breadth_first_search array[0]
puts array.map(&:to_s)