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 JavaScript (judge verdict: Accepted)

//config = { input: 'tmp', newline: '\r\n' }; // win
config = { input: '/dev/stdin', newline: '\n' }; // linux

line = require('fs').readFileSync(config.input, 'ascii')
  .trim()
  .split(config.newline);
n = Number(line.shift());
adj = {};
for (i in line) {
  ary = line[i].split(' ');
  j = ary.shift();
  ary.shift();
  adj[j] = ary;
}

node = new Array(n + 1);
node[0] = null;
node[1] = 0;
for (i = 2; i <= n; i++) node[i] = -1;

d = 1;
do {
  cnt = 0;
  for (i = 1; i <= n; i++) {
    if (node[i] !== d - 1) continue;
    for (j in adj[i]) {
      k = adj[i][j];
      if (node[k] === -1) {
        node[k] = d;
        cnt++;
      }
    }
  }
  d++;
} while (cnt !== 0);

for (i = 1; i <= n; i++)
  console.log("%d %d", i, node[i]);