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 Scala (judge verdict: Accepted)

object Main extends App {
  val n = io.StdIn.readInt()
  val adjacents = Array.fill[Array[Int]](n)(Array())
  for (_ <- 0 until n) {
    val data = io.StdIn.readLine().split(" ").map(_.toInt)
    val id = data(0)-1
    adjacents(id) = data.drop(2).map(_-1)
  }
  val depth = Array.fill[Int](n)(-1)
  def search(x: Int) = {
    val queue = collection.mutable.Queue.empty[Int]
    depth(x) = 0
    queue.enqueue(x)
    loop()
    def loop(): Unit =
      if (queue.isEmpty) ()
      else {
        val i = queue.dequeue()
        for (a <- adjacents(i))
          if (depth(a) < 0) {
            depth(a) = depth(i)+1
            queue.enqueue(a)
          }
        loop()
      }
  }
  search(0)
  for (i <- 0 until n) println(s"${i+1} ${depth(i)}")
}