Go: package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
	"strings"
)

var scanner = bufio.NewScanner(os.Stdin)

func nextString() string {
	scanner.Scan()
	return scanner.Text()
}

func nextInt() int {
	n, err := strconv.Atoi(nextString())
	if err != nil {
		fmt.Printf("strconv.Atoi failed: %v\n", err)
	}
	return n
}

func stringifyArray(arr []int) string {
	return strings.TrimRight(fmt.Sprintf("%+v", arr)[1:], "]")
}

var infty = 1000000000

type nodeKey int

type DirectedGraphNode struct {
	key       nodeKey
	adjacents []*DirectedGraphNode
	visited   bool
	d         int
	parent    *DirectedGraphNode
}

type DirectedGraph struct {
	nodes []DirectedGraphNode
}

func (g *DirectedGraph) setAdjacency(l, r nodeKey) {
	g.nodes[l].adjacents = append(g.nodes[l].adjacents, &g.nodes[r])
}

func newDirectedGraph(nNodes int) *DirectedGraph {
	graph := DirectedGraph{}
	graph.nodes = make([]DirectedGraphNode, nNodes)
	for iNodes := 0; iNodes < nNodes; iNodes++ {
		graph.nodes[iNodes].key = nodeKey(iNodes)

		graph.nodes[iNodes].visited = false
		graph.nodes[iNodes].d = infty
		graph.nodes[iNodes].parent = nil
	}
	return &graph
}

type QueueEntry struct {
	node *DirectedGraphNode
	next *QueueEntry
}

type Queue struct {
	head, tail *QueueEntry
}

func (q *Queue) enqueue(node *DirectedGraphNode) {
	entry := QueueEntry{}
	entry.node = node

	if q.tail == nil {
		q.head = &entry
		q.tail = &entry
		return
	}

	q.tail.next = &entry
	q.tail = q.tail.next
}

func (q *Queue) dequeue() *QueueEntry {
	if q.head == nil {
		return nil
	}
	poped := q.head
	q.head = q.head.next
	if q.head == nil {
		q.tail = nil
	}
	return poped
}

func (q *Queue) isEmpty() bool {
	return q.head == nil
}

func fillDistancesByBreadthFirstSearch(g *DirectedGraph, startNode *DirectedGraphNode) {
	startNode.visited = true
	startNode.d = 0

	queue := Queue{}
	queue.enqueue(startNode)
	for !queue.isEmpty() {
		u := queue.dequeue().node
		for _, adjacent := range u.adjacents {
			if !adjacent.visited {
				adjacent.visited = true
				adjacent.d = u.d + 1
				adjacent.parent = u
				queue.enqueue(adjacent)
			}
		}
		u.visited = true
	}
}

func main() {
	scanner.Split(bufio.ScanWords)

	nNodes := nextInt()
	g := newDirectedGraph(nNodes + 1)

	for iNode := 0; iNode < nNodes; iNode++ {
		nodeKeyLeft := nodeKey(nextInt())
		nAdj := nextInt()
		for iAdj := 0; iAdj < nAdj; iAdj++ {
			nodeKeyRight := nodeKey(nextInt())
			g.setAdjacency(nodeKeyLeft, nodeKeyRight)
		}
	}

	// fmt.Println(g)
	fillDistancesByBreadthFirstSearch(g, &g.nodes[1])
	// fmt.Println(g)

	for iNode := 1; iNode <= nNodes; iNode++ {
		node := g.nodes[iNode]
		distance := node.d
		if distance == infty {
			distance = -1
		}
		fmt.Println(node.key, distance)
	}
}

C++: #include <bits/stdc++.h>
#define rep(i, n) for (ll i = 0; i < (ll)n; i++)
#define PI 3.14159265358979323846
#define vll vector<long long>
#define vvll vector<vector<long long>>
using namespace std;
using ll = long long;
using ld = long double;

// ----本問について----
// 問題：Breadth First Search
// 問題文URL：https://onlinejudge.u-aizu.ac.jp/problems/ALDS1_11_C
// 所要時間：
// 学んだこと：
// ------------------

// ----ここから本問のソースコード----
static const ll N_MAX = 110;

// グローバル変数（入力）
ll n, A[N_MAX][N_MAX];

// 入力
void input() { cin >> n; }

// 幅優先探索
void bfs() {
  queue<ll> q;

  // 距離を-1で初期化
  vll d(n + 1, -1);

  // 始点の処理
  d[1] = 0;
  q.push(1);

  // 探索
  while (!q.empty()) {
    ll v = q.front();
    q.pop();
    rep(i, n) {
      if (A[v][i + 1] == 1 && d[i + 1] == -1) {
        q.push(i + 1);
        d[i + 1] = d[v] + 1;
      }
    }
  }
  rep(i, n) { cout << i + 1 << " " << d[i + 1] << endl; }
}

void solve() {
  // 隣接リストを隣接行列に変換して二次元配列Aに格納
  rep(i, n) {
    ll v, k;
    cin >> v >> k;
    rep(j, k) {
      ll v2;
      cin >> v2;
      A[v][v2] = 1;
    }
  }

  bfs();
}

int main() {
  input();
  solve();
  return 0;
}

C++: #include <iostream>
#include <ios>
#include <vector>
#include <utility>

using namespace std;

void read_data(int n, vector<vector<int>> & Es){
    int u = 0, k = 0, v = 0;
    while (n--){
        cin >> u;
        cin >> k;
        while (k--){
            cin >> v;
            Es[u - 1].push_back(v - 1);
        }
    }
}

void wfs(int node, vector<int> & dist, vector<vector<int>> & Es){
    int d = 0;
    vector<int> frontiers;
    frontiers.push_back(node);
    while (!frontiers.empty()){
        vector<int> new_frontiers;
        for (auto u : frontiers){
            if (dist[u] == -1){
                dist[u] = d;
                for (auto v : Es[u]){
                    new_frontiers.push_back(v);
                }
            }
        }
        frontiers = move(new_frontiers);
        ++d;
    }
}

int main()
{
    ios::sync_with_stdio(false);
    int n = 0;
    cin >> n;
    vector< vector<int> > Es(n);
    vector<int> dist(n, -1);
    read_data(n, Es);
    wfs(0, dist, Es);
    for (int i = 0; i != n; ++i) cout << i + 1 << ' ' << dist[i] << endl;
    
    return 0;
}

C++: #include <iostream>
#include <climits>
using namespace std;

#define N 100
#define MAX 100000

int A[N + 1][N + 1] = {};
int n, u, k, v;
int t1[N + 1]; //発見時刻

//sは始点,tは時間
void bfs(int s, int t) {
    t1[s] = t;
    for(int i = 1; i <= n; ++i) {
        if(A[s][i] == 1 && t < t1[i]) {
            bfs(i, t + 1);
        }
    }
}

int main(void) {
    cin >> n;
    
    for(int i = 0; i < n; ++i) {
        cin >> u >> k;
        for(int j = 0; j < k; ++j) {
            cin >> v;
            A[u][v] = 1;
        }
    }
    
    for(int i = 0; i <= n; ++i) t1[i] = MAX;
    
    bfs(1, 0);
    
    for(int i = 1; i <= n; ++i) {
        cout << i << ' ';
        if(t1[i] == MAX) cout << -1 << endl;
        else cout << t1[i] << endl;
    }
    
    return 0;
}

C++: #include <iostream>
#include <cmath>
#include <vector>
#include <queue>
#include <string>
#include <algorithm>

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int n, u, k, vi;
    std::cin >> n;
    std::vector<std::vector<int>> v(n+1);
    std::queue<int> ex;
    ex.push(1);
    int distance[101]={};
    for (int i=2; i<101; i++) {
        distance[i] = -1;
    }
    for (int i=0; i<n; i++) {
        std::cin >> u >> k;
        for (int j=0; j<k; j++) {
            std::cin >> vi;
            v[u].push_back(vi);
        }
    }

    while (not ex.empty()) {
        int a = ex.front();
        ex.pop();
        for (int i=0; i<v[a].size(); i++) {
            if (distance[v[a][i]] == -1) {
                ex.push(v[a][i]);
                distance[v[a][i]] = distance[a] + 1;
            }
        }
    }

    for (int i=1; i<=n; i++) {
        std::cout << i << " " << distance[i] << "\n";
    }

    return 0;
}