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>

typedef struct T{
  int id;
  int k;
  int *c;
  int parent;
  int depth;
}t;

int main(){

  int i,j,l,n,flag=1,dep=0,par=0;
  t *tree,temp;

  scanf("%d",&n);
  tree = malloc(sizeof(t)*n);
  for(i=0;i<n;i++){
    
    scanf("%d",&tree[i].id);

    scanf("%d",&tree[i].k);
    	tree[i].c = malloc(sizeof(int)*tree[i].k);
     tree[i].depth = 0;
    for(j=0;j<tree[i].k;j++){
      scanf("%d",&tree[i].c[j]);
    }
  }


  /*parent,root????±???????*/

  for(i=0;i<n;i++){
    for(j=0;j<n;j++){
      for(l=0;l<tree[j].k;l++){
	if(tree[i].id==tree[j].c[l]){
	  flag = 0;
	  tree[i].parent = tree[j].id;
	}
      }
    }
    if(flag == 1){
      tree[i].parent = -1;
      flag = 0;
    }
    flag = 1;
  }
  

 /*?????????*/
 for(i=0;i<n;i++){
   flag = 1;
   while(flag){
     flag = 0;
     for(j=n-1;1<=j;j--){
       if(tree[j].id<tree[j-1].id){
	 temp = tree[j-1];
	 tree[j-1] = tree[j];
	 tree[j] = temp;
	 flag = 1;
       }
     }
   }
 }

 /*??±???*/
 for(i=0;i<n;i++){
 	par = tree[i].parent;
      while(par!=-1){
      	dep++;
	  par = tree[par].parent;
	    }
   	 tree[i].depth = dep;
     dep=0;
      }

  for(i=0;i<n;i++){
    printf("node %d: ",tree[i].id);
    printf("parent = %d,",tree[i].parent);

    printf(" depth = %d,",tree[i].depth);

    if(tree[i].k != 0 && tree[i].parent != -1)printf(" internal node,");
    else if(tree[i].parent == -1) printf(" root,");
    else printf(" leaf,");

    if(tree[i].k==0){
	printf(" []\n");
	continue;
      }
    printf(" [");
    for(j=0;j<tree[i].k-1;j++){
      printf("%d, ",tree[i].c[j]);
    }
    printf("%d]",tree[i].c[j]);
    printf("\n");
}

  return 0;
}