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: Time Limit Exceeded)

#include <stdio.h>
#define GRAY  2 
#define WHITE 1
#define BLACK 3 
#define INF -1
int n,color[110],d[110],Q[110],adj[110][110];
int head=0,tail=0;
int judge(){
  if(tail==head) return 1;

  return 0;
}

void enQ(int s){
  Q[tail]=s;
  if(tail==110)tail=0;
  else tail++;
}
int deQ(){
  int a;
  a=Q[head];
  if(head==100)head==0;
  else head++;

  return a;
}

void bfs(int s){
  int i,j,u;
  for(i=2;i<=n;i++){
    color[i]=WHITE;
    d[i]=INF;
  }

  color[s]=GRAY;
  d[s]=0;
 
  enQ(s);

  while(judge() != 1){
 
    u = deQ();
 
    for( j = 1; j <= n; j++ ){
      if(color[j]==WHITE&&adj[u][j]==1){
 
	color[j] = GRAY;
 
	d[j] = d[u] + 1;
 
	enQ(j);
      }
    }
    color[u]=BLACK;
  }

}


int main(){

  int u,k,v,i,j;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d%d",&u,&k);
    for(j=0;j<k;j++){
      scanf("%d",&v);
      adj[u][v]=1;
    }
    bfs(u);
  } 
  bfs(1);

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



}