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

void visit(int);
int n,arr[N+1][N+1],count[N+1],cnt=0;
int main(){
  int i,j,m,node,num;
  
  scanf("%d",&n);
  if(n<1 || n>100) exit(2);
  for(i=0;i<=n;i++){
    for(j=0;j<=n;j++){
      arr[i][j]=0;
    }
  }
  count[0]=999;
  for(i=0;i<n;i++){
    count[i+1]=999;
    scanf("%d %d",&node,&num);
    for(j=0;j<num;j++){
      scanf("%d",&m);
      arr[node][m]=1;
    }
  }
  visit(1);

  for(i=1;i<=n;i++){
    if(count[i] == 999){ // 1の頂点とつながっていないとき
      printf("%d %d\n",i,-1);
    }else{
      printf("%d %d\n",i,count[i]);
    }
  }
  
  return 0;
}

void visit(int x){
  int i;
  count[x]=cnt;
  for(i=1;i<=n;i++){
    if(arr[x][i]==1 && count[i]>cnt){
      cnt++;
      visit(i);
      cnt--;
    }
  }
}