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.Arrays;
import java.util.Queue;
import java.util.Scanner;

public class Main {
	static int n;
	static int[][]M;	// 隣接行列
	static int[]d;
	static final int INF = 10000000;
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		n = scan.nextInt();
		M = new int[n][n];
		d = new int[n];
		for(int i = 0; i < n; i++) {
			int u = scan.nextInt() - 1;
			int k = scan.nextInt();
			for(int j = 0; j < k; j++) {
				M[u][scan.nextInt() - 1] = 1;
			}
		}
		scan.close();
		bfs(0);
	}
	static void bfs(int s) {
		Queue<Integer> queue = new ArrayDeque<Integer>();
		queue.add(s);
		Arrays.fill(d, INF);
		d[s] = 0;
		while(!queue.isEmpty()) {
			int u = queue.poll();
			for(int v = 0; v < n; v++) {
				if(M[u][v] == 0) continue;
				if(d[v] != INF) continue;
				d[v] = d[u] + 1;
				queue.add(v);
			}
		}
		for(int i = 0; i < n; i++) {
			int u = i + 1;
			int l;
			if(d[i] == INF) l = -1;
			else l = d[i];
			System.out.println(u + " " + l);
		}
	}
}