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)

#! /usr/bin/ruby -w
# encoding: utf-8
#

def count_path(node, queue, checked, goal, count)
  next_step = []
  queue.each do |current|
    if current == goal
      return count
    else
      node[current].each do |n|
        if checked[n].nil?
          checked[n] = true
          next_step << n
        end
      end
    end
  end
  if next_step.empty?
    return -1
  else
    return count_path(node, next_step, checked, goal, count + 1)
  end
end

if __FILE__ == $0
  n = gets.to_i
  node = {}
  n.times {
    info = gets.split.map(&:to_i)
    node_num, adj = info[0], info[2..-1]
    node[node_num] = adj
  }

  n.times do |t|
    node_num = t + 1
    puts "#{node_num} #{count_path(node, [1], {}, node_num, 0)}"
  end
end