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++11 (judge verdict: Runtime Error)

#define scanf_s scanf
#include <string>
#include <stdio.h>
#include <math.h>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <iostream>
using namespace std;

#define MAX 100
//#define MAX_ 1000
struct Graph
{
	int id, k;
	list<int> node;
	int d = -1;
};
Graph G[MAX];
//int cou = 0;
void Breadth_cou(Graph *g,int cou)
{
	if (g->k >= 0) {
		for (auto i = g->node.begin(); i != g->node.end(); ++i) {
			if (G[*i].d == -1)
				G[*i].d = cou;
		}
	}
	g->k = 0;
}
void Breadth(Graph *g)
{
	if (g->k >= 0) {
		Breadth_cou(g, g->d + 1);
		for (auto i = g->node.begin(); i != g->node.end(); ++i) {
			Breadth(&G[*i]);
		}
	}
}

int main(void)
{
	int n, u, xk, v;
	scanf_s("%d", &n);
	for (int i = 0; i < n; ++i) {
		scanf_s("%d %d", &u, &xk);
		G[u - 1].id = u - 1;
		G[u - 1].k = xk;
		for (int j = 0; j < xk; ++j) {
			scanf_s("%d", &v);
			G[u - 1].node.push_back(v - 1);
		}
	}
	G[0].d = 0;
	Breadth(&G[0]);
	for (int i = 0; i < n; ++i) {
		printf("%d %d\n", G[i].id + 1, G[i].d);
	}
}