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.*;

public class Main {
    static int WHITE = 0;
    static int GRAY = 1;
    static int BLACK = 2;
    static int[] color;
    static int[] d;
    static int n;
    static int[][] m;

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        n = input.nextInt();
        m = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                m[i][j] = 0;
            }
        }
        for (int i = 0; i < n; i++) {
            int v = input.nextInt();
            int in = input.nextInt();
            for (int j = 0; j < in; j++) {
                int u = input.nextInt();
                m[v - 1][u - 1] = 1;
            }
        }
        bfs();

        for (int i = 0; i < n; i++) {
            System.out.println(i + 1 + " " + d[i]);
        }

    }

    static void bfs() {
        // 初始化
        color = new int[n];
        d = new int[n];
        for (int i = 0; i < n; i++) {
            color[i] = WHITE;
            d[i] = -1;
        }
        color[0] = GRAY;
        d[0] = 0;
        Queue<Integer> Q = new LinkedList<Integer>();
        Q.offer(0); // add()和remove()方法在失败的时候会抛出异常(不推荐)
        while (!Q.isEmpty()) {
            int u = Q.poll();
            for (int v = 0; v < n; v++) {
                if (m[u][v] == 1 && color[v] == WHITE) {
                    color[v] = GRAY;
                    d[v] = d[u] + 1;
                    Q.offer(v);
                }
            }
            color[u] = BLACK;
        }
    }
}