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)

def traverse
  a = []
  length = 1
  now = 0
  @list[now] = 1
  loop do
   @n.times do |i|
     if @m[now][i] == 1 && @list[i] == 0
       a.push([i, length + 1])
       @data[i].length = length
       @list[i] = 1
     end
   end
   if a.empty?
     return nil
   end
   next_one = a.shift
   now = next_one[0]
   length = next_one[1]
  end
  return @list
end

D = Struct.new(:length)
@n = gets.chomp.to_i
@time = 0
@m = Array.new(@n){Array.new(@n){0}}
@data = Array.new(@n){D.new(-1)}
@list = Array.new(@n){0}
@n.times do
	s = gets.split.map(&:to_i)
	i = s[0] - 1
	s[1].times do |l|
		j = s[l + 2] - 1
		@m[i][j] = 1
	end
end
@data[0].length = 0
traverse

@n.times do |i|
	puts "#{ i + 1 } #{@data[i].length}"
end