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 = []
		len = 1
		now = 0
		$list[now] = 1
		loop{
		$n.times do |i|
			if $m[now][i] == 1 && $list[i] == 0
				a << [i, len + 1]
				$data[i].len = len
				$list[i] = 1
			end
		end
	 return nil	if a.empty?
		nex = a.shift
		now = nex[0]
		len = nex[1]
	}
end

D = Struct.new(:len)
$n = gets.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].len = 0
traverse



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