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 Haskell (judge verdict: Accepted)

module Main where

import Debug.Trace
import Data.List
import qualified Data.Sequence as S
import qualified Data.ByteString.Char8 as B
import Data.Int
import Data.Char
import Data.Array
import Data.Maybe
import qualified Data.Map as M
import Data.Ord
import Data.Functor
import Control.Applicative
import Control.Monad
import Text.Printf

solve :: Int -> [[Int]] -> [(Int, Int, Int, String, [Int])]
solve n l = sort (iter 0 (-1) root) where
    iter depth parent node = info:(concatMap (iter (depth+1) node) (children!node)) where
        info :: (Int, Int, Int, String, [Int])
        info = (node, parent, depth,
                case node of
                 _  | node == root      -> "root"
                    | null (children!node)  -> "leaf"
                    | otherwise         -> "internal node",
                    children!node)

    root = iter 0 where
        iter x 
            | tree!x == -1 = x
            | otherwise = iter (tree!x)

    (tree, children) = iter [] [] (sort l) where
        iter ac cc [] = ((listArray (0, n-1) (replicate n (-1))) // ac,
                         (listArray (0, n-1) (replicate n [])) // cc)
        iter ac cc ((x:0:ls):ns) = iter ac cc ns 
        iter ac cc ((x:y:ls):ns) = iter ([(i, x) | i <- ls]++ac) ((x, ls):cc) ns 

readIntLn:: IO Int
readIntLn = (fst . fromJust . B.readInt) <$> B.getLine

readIntList:: IO [Int]
readIntList = map (fst . fromJust . B.readInt) . B.words <$> B.getLine

print' (n, p, d, t, l) = do 
    putStr $ "node "++show n++": parent = "++show p++", depth = "++show d++", "++t++", "
    printInts l

printInts l = do
    putStr "["
    iter l where
        iter [] = putStrLn "]"
        iter [x] = putStrLn $ show x++"]"
        iter (x:xs) = do
            putStr $ show x++", "
            iter xs 

main :: IO() 
main = do
    n <- readIntLn
    l <- replicateM n readIntList
    mapM_ print' $ solve n l