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)

#
# ???????????¢?´¢
#
n = gets.chomp.to_i
graph = [[]]
n.times do
  graph << gets.chomp.split.map { |s| s.to_i }
end

dis = []
que = []
visited = []

dis[1] = 0
2.upto(n) do |i|
  dis[i] = -1
end
que << 1

# debug
# p graph
until que.empty?
  current_node = que.shift
  visited << current_node
  graph[current_node][1].times do |i|
    next_node = graph[current_node][2+i]
    if dis[next_node] == -1
      dis[next_node] = dis[current_node] + 1
    end
    
    if !(visited.include?(next_node)) and !(que.include?(next_node))
      que << next_node
    end
    # debug
    # p que
  end
end

1.upto(n) do |i|
  puts "#{i} #{dis[i]}"
end