Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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.*;

public class Main{

	static int[] list;
	static int pp = 0, pn = 1; // pp?§?????????????????????????????§???????????????????????????????pn

	public static void main(String[] args){

		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int[][] v = new int[n+1][n+1];
		int[] d = new int[n+1];

		for(int i = 0; i < n; i++){
			int u = scan.nextInt();
			int k = scan.nextInt();
			d[u] = -1;
			for(int j = 1; j <= k ; j++)
				v[u][j] = scan.nextInt();
		}

		d[1] = 0; // ???1??????????????¢???0
		list = new int[n+1];
		list[0] = 1; // ?§????????????????1??????

		Bsearch(v, d);

		for(int i = 1; i <= n; i++)
			System.out.println(i + " " + d[i]);

		scan.close();
		System.exit(0);
	}

	private static void Bsearch(int[][] v, int[] d){
		for(; pp < pn; pp++){
			int fr = list[pp]; // fr????§????????§???????????????????????????????
			for(int i = 1; i < v[fr].length-1; i++){
				int to = v[fr][i]; // ?§????????????????????????????
				if(d[to] == -1){ // ???????????§????????¢???-1??????
					d[to] = d[fr] + 1; // ??????????????¢????§?????????§????????¢+1
					list[pn] = to; // ?§?????????????????????????????´?
					pn++; // ???????????????????????????+1
					// 2??????1??????????????¨???list[pn++] = to??????????????????????????????pn???+1?????????????????????????????????
				}
			}
		}
	}
}