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>

struct tree{
    int id;
    int parent;
    int children[20];
    int child;
    int depth;
};

struct tree ki[100000];
void bubbleSort(int); 

int main(){
    int key,i,j,k,n,temp;
    scanf("%d",&n);
 
    for(i = 0 ; i < n ; i++){
        scanf("%d %d",&ki[i].id,&ki[i].child);
        if(ki[i].child > 0){
            for(j = 0 ; j < ki[i].child ; j++){
                scanf("%d",&ki[i].children[j]);
            }
        }
    }
   
    bubbleSort(n);

    for(i = 0 ; i < n ; i++){
        key = ki[i].id;
        ki[i].parent = -1;
        for(j = 0 ; j < n ; j++){
            if(i!=j){
                if(ki[j].child > 0){
                    for(k = 0 ; k < ki[j].child ; k++){
                        if(key == ki[j].children[k]){
                            ki[i].parent = ki[j].id;
                        }
                    }
                }
            }
        }
        if(ki[i].parent == -1){
            ki[i].depth = 0;
            temp = i;
        }
    }

    if(ki[temp].child > 0){
        for(i = 0 ; i < ki[temp].child ; i++){
            ki[ki[temp].children[i]].depth = 1;
        }
        for(i = 0 ; i < 21 ; i++){
            for(j = 0 ; j < n ; j++){
                if(j!=temp){
                    ki[j].depth = ki[ki[j].parent].depth + 1; 
                }               
            }
        }
    }


    for(i = 0 ; i < n ; i++){
        printf("node %d: parent = %d, depth = %d, ",i,ki[i].parent,ki[i].depth);
        if(ki[i].child > 0 || ki[i].depth == 0){
            if(ki[i].depth == 0) printf("root, [");
            else printf("internal node, [");
            for(j = 0 ; j < ki[i].child ; j++){
                printf("%d",ki[i].children[j]);
                if(j != ki[i].child - 1) printf(", ");
            }
            printf("]\n");
        }
        else printf("leaf, []\n");
    }

    return 0;
            
}
void bubbleSort(int len){
    int temp,temp2,i,j,k;

    for(i = 0 ; i < len ; i++){

        for(j = len-1 ; j > i ; j--){
        
            if(ki[j].id < ki[j-1].id){
        
                temp = ki[j].id;
                ki[j].id = ki[j-1].id;
                ki[j-1].id = temp;
                temp = ki[j].child;
                ki[j].child = ki[j-1].child;
                ki[j-1].child = temp;
        
                if(temp > 0 && temp > ki[j].child){
                    for(k = ki[j].child ; k < 20 ; k++){
                        ki[j-1].children[k] = 0;
                    }             
                    for(k = 0 ; k < temp ; k++){
                        temp2 = ki[j].children[k];
                        ki[j].children[k] = ki[j-1].children[k];
                        ki[j-1].children[k] = temp2;
                    }
                }
                else if(ki[j].child > 0 && ki[j].child > temp){
                    for(k = temp ; k < 20 ; k++){
                        ki[j].children[k] = 0;
                    }             
                    for(k = 0 ; k < ki[j].child ; k++){
                        temp2 = ki[j-1].children[k];
                        ki[j-1].children[k] = ki[j].children[k];
                        ki[j].children[k] = temp2;
                    }

                }
            }
        }
    }
}