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.to_i
a = N.times.map{gets.split.map(&:to_i)}

INFTY = 1<<21
n = 100

@m = Array.new(n){Array.new(n){0}}
@d = Array.new(N){INFTY}

def bfs(s)
  q = []
  q.push(s)

  @d[s] = 0

  while !q.empty?
    u = q.shift

    (0...N).each do |v|
      next if @m[u][v] == 0
      next if @d[v] != INFTY

      @d[v] = @d[u] + 1
      q.push(v)
    end
  end

  (0...N).each do |i|
    puts "#{i + 1} #{@d[i] == INFTY ? -1 : @d[i]}"
  end
end

(0...N).each do |i|
  u, k = a[i][0], a[i][1]
  u -= 1

  (0...k).each do |j|
    v = a[i][j + 2]
    v -= 1

    @m[u][v] = 1
  end
end
bfs(0)