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 C (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

#define N 100

int n, l, j, u, k, tmp;

int list[N][N];
int lengthlist[N];

void search(int , int );

int main(){
  
  scanf("%d", &n);
  
  for(l = 0 ; l < n ; l++){
    scanf("%d", &u);
    scanf("%d", &k);
    
    for( j = 0 ; j < k ; j++){
      scanf("%d", &tmp);
      list[u-1][tmp-1] = 1;
    }
  }
  
  for( l = 0 ; l < n ; l++){
    lengthlist[l] = -1;
  }
  
  search(0, -1);

  for(j = 0 ; j < n ; j++){
    printf("%d %d\n", j+1, lengthlist[j]);
  }

  return 0;
}

void search(int y, int length){
  int i;
  
  length++;
  lengthlist[y] = length;


  for( i = 0 ; i < n ; i++){
    if( list[y][i] == 1 ){
      if( lengthlist[i] <= length && lengthlist[i] >= 0 ){
        continue;
      }
      search(i, length);
    }
  }
}