Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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 N 100
#define MAX 100000000
#define WHITE 0
#define GRAY 1
#define BLACK 2

int n, tail, head;
int Q[N], color[N], d[N], M[N][N];

void enqueue(int x){

    Q[tail] = x;

  if(tail+1 == N){
    tail = 0;
  } else {
    tail++;
  }
}

int dequeue(){
  int x;

  x = Q[head];
  if(head+1 == N){
    head = 0;
  } else {
    head++;
  }
  return x;
}

void bfs(int s){

  int i, j;

  for(i=0; i<n; i++){
    color[i] = WHITE;
    d[i] = MAX;
  }
  color[s] = GRAY;
  d[s] = 0;

  enqueue(s);

  while(Q!=0){
    i = dequeue();
    for(j=0; j<n; j++){
      if(color[j] = WHITE){
        color[j] = GRAY;
        d[j] = d[i]+1;
        enqueue(j);
      }
    }
    color[i] = BLACK;
  }
}

int main(){

  int i, j, n, u, k, v;

  scanf("%d", &n);
  for(i=0; i<n; i++){
    for(j=0; j<n; j++){
      M[i][j] = 0;
    }
  }

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

  bfs(0);

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

  return 0;
}