Go: package main

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

/***************************************************************
stackによる深さ優先探索を実装する。
***************************************************************/

const (
	N     = 100
	WHITE = 0
	GRAY  = 1
	BLACK = 2
)

var (
	n            int
	Adj          [N][N]int
	statusColor  [N]int
	discoverTime [N]int
	finishTime   [N]int
	clock        int
	nt           [N]int
)

/***************************************************************
Stackの定義:
- field
	container: ノードの格納されたスライス
	sp: スタックポインタ
- method
	size(): スタックに積まれているノードの個数を返す
	top(): スタックの一番上に積まれているノードを返す
	isEmpty(): スタックが空かどうかを返す
	pop(): スタックの要素をポップし削除する
	push(): スタックに要素を積む
***************************************************************/

type Stack struct {
	container []int
	sp        int
}

func (s *Stack) size() int {
	return len(s.container)
}

func (s *Stack) top() int {
	return s.container[s.sp]
}

func (s *Stack) isEmpty() bool {
	return s.sp == 0
}

func (s *Stack) pop() int {
	_pop := func() {
		s.container = s.container[:len(s.container)-1]
	}
	defer _pop()
	if s.isEmpty() {
		panic("There is no elements in stack!")
	}
	s.sp--
	return s.container[s.sp]
}

func (s *Stack) push(n int) {
	s.sp++
	s.container[s.sp] = n
}

func next(u int) int {
	for v := nt[u]; v < n; v++ {
		nt[u] = v + 1
		if Adj[u][v] == 1 {
			return v
		}
	}
	return -1
}

func _DFS(target int) {
	for i := 0; i < n; i++ {
		nt[i] = 0
	}
	stack := new(Stack)
	stack.container = make([]int, N)
	stack.push(target)
	statusColor[target] = GRAY
	clock++
	discoverTime[target] = clock
	for !stack.isEmpty() {
		u := stack.top()
		v := next(u)
		if v != -1 {
			if statusColor[v] == WHITE {
				statusColor[v] = GRAY
				clock++
				discoverTime[v] = clock
				stack.push(v)
			}
		} else {
			stack.pop()
			statusColor[u] = BLACK
			clock++
			finishTime[u] = clock
		}
	}
}

func DFS() {
	for i := 0; i < n; i++ {
		statusColor[i] = WHITE
		nt[i] = 0
	}
	clock = 0
	for i := 0; i < n; i++ {
		if statusColor[i] == WHITE {
			_DFS(i)
		}
	}
	for i := 0; i < n; i++ {
		fmt.Printf("%d %d %d\n", i+1, discoverTime[i], finishTime[i])
	}
}

func main() {
	scanner := bufio.NewScanner(os.Stdin)
	scanner.Split(bufio.ScanWords)
	scanner.Scan()
	n, _ = strconv.Atoi(scanner.Text())
	for i := 0; i < n; i++ {
		for j := 0; j < n; j++ {
			Adj[i][j] = 0
		}
	}
	for i := 0; i < n; i++ {
		scanner.Scan()
		u, _ := strconv.Atoi(scanner.Text())
		scanner.Scan()
		k, _ := strconv.Atoi(scanner.Text())
		u--
		for j := 0; j < k; j++ {
			scanner.Scan()
			v, _ := strconv.Atoi(scanner.Text())
			v--
			Adj[u][v] = 1
		}
	}

	DFS()
}

C++: #include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <climits>
#include <cstring>


using namespace std;

const int N_MAX = 1000000;

int main(){
  int n;
  cin >> n;
  stack<pair<int, int> > edges;
  stack<pair<int, int> > paths;
  
  int** adjl = new int*[n];
  int* d = new int[n];
  int* f = new int[n];
  
  for(int i=0; i<n; i++){
    adjl[i] = new int[n];
    memset(adjl[i], -1, n * sizeof(adjl[i][0]));
  }
  
  memset(d, -1, n*sizeof(d[0]));
  memset(f, -1, n*sizeof(f[0]));

  for(int i=0; i<n; i++){
    int row_num, k;
    cin >> row_num >> k; row_num--;
    
    for(int j=0; j<k; j++){
      int node_num;
      cin >> node_num; node_num--;
      adjl[row_num][j] = node_num; // [[3, 5, -1 , ...], [1, 3, -1, ...], [1, 2], ...]
    }
  }

  for(int row_num=n-1; row_num>=0; row_num--){
      pair<int, int> p;
      p.first = -1; p.second = row_num; // visited from nowhere
      edges.push(p);
    }

  int time = 1;
  while(!edges.empty()){
    //    cout << "time: " << time << endl;
    
    pair<int, int> edge;
    edge = edges.top(); edges.pop();
    int st, en; st = edge.first; en = edge.second;
    //    cout << "visiting: " << en << " from: " << st << endl;
    
    pair<int, int> path; 
    // record path to go back
    if(st > -1){ // exclude (-1, node)
      path.first = en; path.second = st;
      // avoid inf loop: if the node is not finished, have to go back at some point
      if(f[path.second] == -1){
	paths.push(path);
	// cout << "<" << path.first << ", " << path.second << "> pushed as a path." << endl;
      }
    }    
    
    // if first time visit, mark as visited
    if(d[en] == -1) d[en] = time;
    
    // if not first time, check if it was finished.
    // if it was already finished, continue
    if(f[en] != -1){
      continue;
    }
    
    // counter for opened nodes
    int unvisited_node_num = 0;

    // open node (reverse order)
    for(int i=n-1; i>=0; i--){
      int next_node = adjl[en][i];
      // cout << "next_node: " << next_node << endl;
      // undefined dest or visited, continue
      if(next_node == -1) continue;
      if(d[next_node] != -1){
	// if num_to_visit[en] > 0;
	continue;
      }
      // if unvisited, push into stack
      unvisited_node_num++;
      pair<int, int> next_edge; next_edge.first = en; next_edge.second = next_node;
      edges.push(next_edge);
      // cout << "<" << next_edge.first << ", " << next_edge.second << "> pushed." << endl;
    }

    // if there are no unvisited linked nodes anymore, mark the node finished
    if(unvisited_node_num == 0){
      // cout << "going back..." << endl;
      f[en] = time + 1;
      // cout << "f[" << en << "]: " << f[en] << endl;
      
      // way to go back
      if(paths.empty()) break;
      path = paths.top(); paths.pop();
      edges.push(path);
      // cout << "path pushed." << endl;
    }
    time++;
  }

  // print 'i+1, d[i], f[i]'
  
  for(int i=0; i<n; i++){
    cout << i+1 << " " << d[i] << " " << f[i] << endl;
  }
  
  for(int i=0; i<n; i++){
    delete[] adjl[i];
  }
  delete[] adjl;
  delete[] d;
  delete[] f;
  
  return 0;
}

C++: #include <bits/stdc++.h>
#define REP(i,n) for (int i = 0; i <(n); ++i)
#define ALL(v) v.begin(), v.end()
using namespace std;
//using ll = long long;
//using P = pair<int,int>;
//static const double PI = acos(-1);


//---------------------------//
//DFS 再帰関数を利用


static const int N     = 100;
static const int WHITE = 0;
static const int GRAY  = 1;
static const int BLACK = 2;

//隣接行列
int n, M[N][N];
int color[N], d[N], f[N], tt;

void dfs_visit(int u){
  //cout << "u: " << u << endl; //来訪地点のチェック
  color[u] = GRAY;
  d[u] = ++tt;  //最初に訪れたときの処理
  REP(v,n){
    if(M[u][v] == 0) continue;
    //cout << "v: " << v << endl; // 次に向かう点をチェック
    if(color[v] == WHITE) {
      dfs_visit(v); //再帰
    }
  }
  color[u] = BLACK;
  f[u] = ++tt;  //最後に訪れたときの処理

}


void dfs(){
  //int u;
  // 初期化
  REP(u,n) color[u] = WHITE;
  tt = 0;
  //
  REP(u,n) {
    if( color[u] == WHITE) dfs_visit(u); //再帰関数に
  }
  REP(u,n){
    cout << u+1 << " " << d[u] << " " << f[u] << endl;
    //cout << color[u] << endl;
  }


}


int main(){
  int u,k,v;
  cin >> n;
  //隣接行列の初期化
  REP(i,n) REP(j,n) M[i][j] = 0;


  //隣接行列の作製
  REP(i,n){
    cin >> u >> k;
    --u;
    REP(j,k){
      cin >> v;
      --v;
      M[u][v] = 1;
    }

  }
  //dfs
  dfs();


  //隣接行列のoutput(確認用)
  //REP(i,n) {REP(j,n) cout << M[i][j];cout << endl;}



  return 0;
}