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)

# Node
class Node
  attr_accessor :id, :edges, :visited, :cost

  def initialize(id, edges = [], visited = false, cost = nil)
    @id, @edges, @visited, @cost = id, edges, visited, cost
  end
end

# Edge
class Edge
  attr_accessor :to

  def initialize(to)
    @to = to
  end
end

# Graph
class Graph
  attr_accessor :nodes

  def initialize(data)
    @nodes = data.map do |id, edges|
      edges.map! { |edge| Edge.new(*edge) }
      Node.new(id, edges)
    end
  end

  def bfs(start)
    @nodes.each do |node|
      if node.id == start
        node.visited = true
        node.cost = 0
      else
        node.visited = false
        node.cost = -1
      end
    end

    queue = []
    queue << @nodes.find { |node| node.id == start }
    # queue = [start]

    # count = 0
    while queue.any?
      node = queue.shift
      node.visited = true
      node.edges.each do |edge|
        id = edge.to
        visited_node = @nodes.find { |item| item.id == id }
        unless visited_node.visited
          visited_node.cost = node.cost + 1 if visited_node.cost == -1
          queue << visited_node
        end
      end
    end
  end
end

n = gets.chop.to_i

data = {}

n.times do
  u, k, *v = gets.chop.split.map(&:to_i)
  data[u] = v
end

g = Graph.new data
g.bfs(1)

g.nodes.each do |node|
  puts "#{node.id} #{(node.visited) ? node.cost : -1}"
end