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++14 (judge verdict: Runtime Error)

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<sstream>
#include<cmath>
#include<numeric>
#include<map>
#include<stack>
#include<queue>
using namespace std;
int inf = 1000000000;

class Graph{
public:
  int v;
  int cost = inf;
  int ftime = -1;
  int ltime = -1;
  vector<int> e;
};
vector<Graph> g;
int cnt = 0;
void dfs(int pos){
  if( g[pos].ftime != -1 ) return;
  cnt++;
  g[pos].ftime = cnt;
  for(int i=0; i<g[pos].e.size(); i++){
    dfs(g[pos].e[i]);
  }
  cnt++;
  g[pos].ltime = cnt;
}
void bfs(int pos){
  for(int i=0; i<g[pos].e.size(); i++){
    // if( g[ g[pos].e[i] ].cost != inf ) continue;
    g[ g[pos].e[i] ].cost = min(g[ g[pos].e[i] ].cost, g[pos].cost + 1);
    bfs(g[pos].e[i]);
  }
}

int main(void) {

  int n;
  cin >> n;
  g.resize(n);
  int u, k;
  for(int i=0; i<n; i++){
    cin >> u >> k;
    u--;
    g[u].v = u;
    g[u].e.resize(k);
    for(int j=0; j<k; j++){
      cin >> g[u].e[j];
      g[u].e[j]--;
    }
    sort(g[u].e.begin(), g[u].e.end());
  }

  g[0].cost = 0;
  bfs(0);
  for(int i=0; i<n; i++) cout << g[i].v + 1 << " " << (g[i].cost == inf ? -1 : g[i].cost) << endl;

  // for(int i=0; i<n; i++) dfs(i);
  // for(int i=0; i<n; i++){
  //   cout << g[i].v + 1 << " " << g[i].ftime << " " << g[i].ltime << endl;
  // }



  return 0;
}

// EOF