Problem:
Rooted Trees

A graph G = (V, E) is a data structure where V is a finite set of vertices and E is a binary relation on V represented by a set of edges. Fig. 1 illustrates an example of a graph (or graphs).

Fig. 1

A free tree is a connnected, acyclic, undirected graph. A rooted tree is a free tree in which one of the vertices is distinguished from the others. A vertex of a rooted tree is called "node."

Your task is to write a program which reports the following information for each node u of a given rooted tree T:

node ID of u

parent of u

depth of u

node type (root, internal node or leaf)

a list of chidlren of u

If the last edge on the path from the root r of a tree T to a node x is (p, x), then p is the parent of x, and x is a child of p. The root is the only node in T with no parent.

A node with no children is an external node or leaf. A nonleaf node is an internal node

The number of children of a node x in a rooted tree T is called the degree of x.

The length of the path from the root r to a node x is the depth of x in T.

Here, the given tree consists of n nodes and evey node has a unique ID from 0 to n-1.

Fig. 2 shows an example of rooted trees where ID of each node is indicated by a number in a circle (node). The example corresponds to the first sample input.

Fig. 2

Input

The first line of the input includes an integer n, the number of nodes of the tree.

In the next n lines, the information of each node u is given in the following format:

id k c1 c2 ... ck

where id is the node ID of u, k is the degree of u, c1 ... ck are node IDs of 1st, ... kth child of u. If the node does not have a child, the k is 0.

Output

Print the information of each node in the following format ordered by IDs:

node id: parent = p, depth = d, type, [c1...ck]

p is ID of its parent. If the node does not have a parent, print -1.

d is depth of the node.

type is a type of nodes represented by a string (root, internal node or leaf). If the root can be considered as a leaf or an internal node, print root.

c1...ck is the list of children as a ordered tree.

Please follow the format presented in a sample output below.

Constraints

1 ≤ n ≤ 100000

Sample Input 1

13
0 3 1 4 10
1 2 2 3
2 0
3 0
4 3 5 6 7
5 0
6 0
7 2 8 9
8 0
9 0
10 2 11 12
11 0
12 0

Sample Output 1

node 0: parent = -1, depth = 0, root, [1, 4, 10]
node 1: parent = 0, depth = 1, internal node, [2, 3]
node 2: parent = 1, depth = 2, leaf, []
node 3: parent = 1, depth = 2, leaf, []
node 4: parent = 0, depth = 1, internal node, [5, 6, 7]
node 5: parent = 4, depth = 2, leaf, []
node 6: parent = 4, depth = 2, leaf, []
node 7: parent = 4, depth = 2, internal node, [8, 9]
node 8: parent = 7, depth = 3, leaf, []
node 9: parent = 7, depth = 3, leaf, []
node 10: parent = 0, depth = 1, internal node, [11, 12]
node 11: parent = 10, depth = 2, leaf, []
node 12: parent = 10, depth = 2, leaf, []

Sample Input 2

4
1 3 3 2 0
0 0
3 0
2 0

Sample Output 2

node 0: parent = 1, depth = 1, leaf, []
node 1: parent = -1, depth = 0, root, [3, 2, 0]
node 2: parent = 1, depth = 1, leaf, []
node 3: parent = 1, depth = 1, leaf, []

Note

You can use a left-child, right-sibling representation to implement a tree which has the following data:

the parent of u

the leftmost child of u

the immediate right sibling of u

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>
#include <stdlib.h>
#include <string.h>

#define ROOT 0
#define INTERNAL 1
#define LEAF 2

struct node {
  int id;
  int k;
  struct node *parent;
  int depth;
  int type;
  int *child_id;
  struct node **child;
};

struct node *tree;
int n;

struct node *search_node(int id);
void set_parameter(struct node *p);
void quicksort(struct node *A, int p, int r);
int partition(struct node *A, int p, int r);

int main(int argc, char *argv[])
{
  int i, j;
  struct node *tmp;
  
  scanf("%d", &n);

  tree = (struct node *)malloc(sizeof(struct node) * n);

  for (i = 0; i < n; i++) {
    scanf("%d", &(tree[i].id));
    scanf("%d", &(tree[i].k));
    if (tree[i].k != 0) {
      tree[i].child_id = (int *)malloc(sizeof(int) * (tree[i].k));
      tree[i].child = (struct node **)malloc(sizeof(struct node *) * (tree[i].k));
      for (j = 0; j < tree[i].k; j++) {
	scanf("%d", &(tree[i].child_id[j]));
      }
    } else {
      tree[i].child_id = NULL;
      tree[i].child = NULL;
    }
    tree[i].parent = NULL;
  }

  quicksort(tree, 0, n - 1);

  for (i = 0; i < n; i++) {
    for (j = 0; j < tree[i].k; j++) {
      tmp = search_node(tree[i].child_id[j]);
      if (tmp == NULL) {
	fprintf(stderr, "error\n");
	exit(0);
      }
      tmp->parent = &tree[i];
      tree[i].child[j] = tmp;
    }
  }

  for (i = 0; i < n; i++) {
    if (tree[i].parent == NULL) break;
  }
  set_parameter(&tree[i]);

  for (i = 0; i < n; i++) {
    printf("node %d: ", tree[i].id);
    
    if (tree[i].parent == NULL) printf("parent = -1, ");
    else printf("parent = %d, ", tree[i].parent->id);
    
    printf("depth = %d, ", tree[i].depth);
    
    if (tree[i].type == ROOT) printf("root, ");
    else if (tree[i].type == INTERNAL) printf("internal node, ");
    else printf("leaf, ");

    printf("[");
    for (j = 0; j < tree[i].k; j++) {
      if (j != tree[i].k - 1) printf("%d, ", tree[i].child[j]->id);
      else printf("%d", tree[i].child[j]->id);
    }
    printf("]\n");
  }

  return 0;
}

struct node *search_node(int id)
{
  int i;

  for (i = 0; i < n; i++) {
    if (tree[i].id == id) return &tree[i];
  }
  return NULL;
}

void set_parameter(struct node *p)
{
  int i;

  if (p->parent == NULL) {
    p->depth = 0;
    p->type = ROOT;
  } else {
    p->depth = p->parent->depth + 1;
    if (p->k == 0) p->type = LEAF;
    else p->type = INTERNAL;
  }

  for (i = 0; i < p->k; i++) set_parameter(p->child[i]);
}

void quicksort(struct node *A, int p, int r)
{
  int q;
  
  if (p < r) {
    q = partition(A, p, r);
    quicksort(A, p, q-1);
    quicksort(A, q+1, r);
  }
}

int partition(struct node *A, int p, int r)
{
  int x = A[r].id, i = p - 1, j;
  struct node tmp;

  for (j = p; j < r; j++) {
    if (A[j].id <= x) {
      i++;
      memcpy(&tmp, &A[j], sizeof(struct node));
      memcpy(&A[j], &A[i], sizeof(struct node));
      memcpy(&A[i], &tmp, sizeof(struct node));
    }
  }
  memcpy(&tmp, &A[r], sizeof(struct node));
  memcpy(&A[r], &A[i+1], sizeof(struct node));
  memcpy(&A[i+1], &tmp, sizeof(struct node));

  return i+1;
}