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.Linq;
using System.Collections.Generic;
using static System.Console;
using System.Text;
using System.IO;

namespace AOJ
{
    class Program
    {
        static public long[] Sarray() { return ReadLine().Split().Select(long.Parse).ToArray(); }
        static public List<long> Slist() { return ReadLine().Split().Select(long.Parse).ToList(); }

        static void Main(string[] args)
        {
            var N = Sarray()[0];
            var ringyou = new long[N + 1, N + 1];
            for (var i = 1; i <= N; ++i)
            {
                var ukv = Sarray();
                for (var j = 2; j < ukv.Length; ++j)
                {
                    ringyou[i, ukv[j]] = 1;
                }
            }
            var d = new long[N + 1];
            for (var i = 0; i < N + 1; ++i) d[i] = -1;
            var q = new Queue<long>();
            q.Enqueue(1);
            d[1] = 0;
            while(q.Any())
            {
                var v = q.Dequeue();
                for (var i = 1; i < N + 1; ++i)
                {
                    if (0 != ringyou[v, i] && d[i] == -1)
                    {
                        q.Enqueue(i);
                        d[i] = d[v] + 1;
                    }
                }
            }
            for (var i = 1; i <= N; ++i)
            {
                WriteLine($"{i} {d[i]}");
            }
        }
    }
}