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: Wrong Answer)

def fromAdjRow(xs, n)
  a = []
  1.upto(n) do |i|
    a << i if xs[i-1] == 1
  end
  a
end

def toAdj(g, n)
  g.map do |xs|
    a = [0]*n
    unless xs[1] == 0
      xs[2..-1].each do |x|
        a[x-1] = 1
      end
    end
    a
  end
end

if __FILE__ == $0
  n = gets.to_i
  g = toAdj(n.times.map{ gets.split.map(&:to_i) }, n)

  q = [1]
  visited = Array.new(n, false)
  visited[0] = true
  ds = Array.new(n, -1)
  ds[0] = 0
  d  = 1
  until q.empty?
    u  = q.pop
    fromAdjRow(g[u-1], n).each do |v|
      unless visited[v-1]
        ds[v-1] = d
        visited[v-1] = true
        q.push(v)
      end
    end
    d += 1
  end

  ds.each.with_index{ |x, i| puts (i+1).to_s + " " + x.to_s }
end