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++14 (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;

std::vector<int> v[100005];
struct Data
{
    int parent_index, depth;
    std::string type;
};
std::map<int, Data> data;
void dfs(const int index, const int parent_index, const int depth)
{
    int size = v[index].size();
    
    Data d;
    d.parent_index = parent_index;
    d.depth = depth;
    if(d.parent_index == -1) d.type = "root";
    else if(size == 1) d.type = "leaf";  // 接点が親だけ
    else d.type = "internal node";
    data[index] = d;
    
    for(int i=0; i<size; i++){
        int child = v[index][i];
        if(child == parent_index) continue;
        dfs(child, index, depth+1);
    }
}

int main()
{
    int n;
    cin >> n;
    std::vector<int> childlist;
    for(int i=0; i<n; i++){
        int id, k, c;
        cin >> id >> k;
        for(int j=0; j<k; j++){
            cin >> c;
            v[id].push_back(c);
            v[c].push_back(id);
            childlist.push_back(c);  // root検索用
        }
    }
    int root = -1;
    for(int i=0; i<n; i++){
        std::vector<int>::iterator itr;
        itr = std::find(childlist.begin(), childlist.end(), i);
        if(itr == childlist.end()){
            root = i;
        }
    }
    if(root == -1) cout << "root not found." << endl;
    dfs(root, -1, 0);
    
    for(int index=0; index<n; index++){
        
        Data d = data[index];
        cout << "node " << index << ": parent = " << d.parent_index
            << ", depth = " << d.depth << ", " << d.type << ", [";
        
        int size = v[index].size();
        if(index != root){
            std::vector<int>::iterator itr;
            itr = find(v[index].begin(), v[index].end(), d.parent_index);
            v[index].erase(itr); size--;
        }
        for(int i=0; i<size; i++){
            int child = v[index][i];
            //if(child == d.parent_index) continue;
            cout << child;
            if(i != size-1) cout << ", ";
        }
        cout << "]" << endl;
    }
}