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.
题意：
输出给定有根树的T中的各结点u的信息，信息内容如下。
- u的结点编号
- u的结点种类（根、内部结点、叶）
- u的父结点编号
- u的子结点列表
- u的深度
输入：第一行是结点的个数n。接下来的n行按照下述格式输入各结点的信息，每个结点占一行。
id k c1c2...ck
id为结点编号 k为度。c1c2...ck为第一个子结点到第k个子结点的编号。
输出：请看输出示例。
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1e5+5;
struct node
{
int p,l,r;//左子右兄弟表示法，l代表节点u的最左侧的子结点，r为u的右侧紧邻的兄弟节点
};
node T[maxn];
int n,D[maxn];
void print(int u)
{
cout<<"node "<<u<<": ";
cout<<"parent = "<<T[u].p<<", ";
cout<<"depth = "<<D[u]<<", ";
if(T[u].p==-1)//不存在父结点，即为根节点
cout<<"root, ";
else if(T[u].l==-1)//没有子结点，即为叶
cout<<"leaf, ";
else
cout<<"internal node, ";
cout<<"[";
for(int i=0,c=T[u].l;c!=-1;i++,c=T[c].r)
{
if(i)
cout<<", ";
cout<<c;//节点u的子结点列表从u的左侧子结点开始按顺序输出，直到当前子结点不存在右侧兄弟节点为止
}
cout<<"]"<<endl;
}
void rec(int u,int p)//递归求结点的深度
{
D[u]=p;
if(T[u].r!=-1)//当前结点存在右侧兄弟节点，不改变深度
rec(T[u].r,p);
if(T[u].l!=-1)//存在最左侧子结点，深度+1
rec(T[u].l,p+1);
}
int main()
{
cin>>n;
for(int i=0;i<n;i++)
T[i].p=T[i].l=T[i].r=-1;//初始化
for(int i=0;i<n;i++)
{
int u,k;
cin>>u>>k;
for(int j=0;j<k;j++)
{
int c,l;
cin>>c;
if(j)
T[l].r=c;
else
T[u].l=c;
l=c;
T[c].p=u;
}
}
int r;//根节点的编号
for(int i=0;i<n;i++)
if(T[i].p==-1)
r=i;
rec(r,0);
for(int i=0;i<n;i++)
print(i);
return 0;
}