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 JAVA (judge verdict: Accepted)

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.next());
        Map<Integer, List<Integer>> graph = new HashMap<>();
        for(int i = 0; i < n; i++){
            int u = Integer.parseInt(sc.next());
            graph.put(u, new ArrayList<>());
            int k = Integer.parseInt(sc.next());
            for(int j = 0; j < k; j++){
                graph.get(u).add(Integer.parseInt(sc.next()));
            }
        }
        
        ArrayDeque<Integer> queue = new ArrayDeque<>();
        int[] d = new int[n+1];
        Arrays.fill(d, -1);
        queue.add(1);
        d[1] = 0;
        while(!queue.isEmpty()){
            int u = queue.poll();
            graph.get(u).forEach((v) -> {
                if(d[v] == -1){
                    queue.add(v);
                    d[v] = d[u] +1;
                }
            });
        }
        
        StringBuilder sb = new StringBuilder();
        for(int i = 1; i <= n; i++){
            sb.append(i).append(" ").append(d[i]).append(System.lineSeparator());
        }
        System.out.print(sb);
    }
}