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 C# (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
    class Node
    {
        internal int d;
        internal List<int> adj;

        public Node()
        {
            d = -1;
            adj = new List<int>();
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());

            Node[] nodes = new Node[n + 1];

            for (int i = 0; i < n; i++)
            {
                int[] input = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);

                nodes[input[0]] = new Node();

                for (int j = 0; j < input[1]; j++)
                {
                    nodes[input[0]].adj.Add(input[2 + j]);
                }
            }

            nodes[1].d = 0;

            Queue<Node> Q = new Queue<Node>();
            Q.Enqueue(nodes[1]);

            while (Q.Count > 0)
            {
                Node temp = Q.Dequeue();

                for (int i = 0; i < temp.adj.Count; i++)
                {
                    if (nodes[temp.adj[i]].d == -1)
                    {
                        nodes[temp.adj[i]].d = temp.d + 1;
                        Q.Enqueue(nodes[temp.adj[i]]);
                    }
                }
            }

            StringBuilder sb = new StringBuilder();
            for (int i = 1; i <= n; i++)
            {
                sb.AppendLine(i + " " + nodes[i].d);
            }
            Console.Write(sb);
        }
    }
}