【有根树】Rooted Trees C++

题目来源：Aizu - ALDS1_7_A

题目：

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 ... c_k

where id is the node ID of u, k is the degree of u, c₁ ... c_k 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...c_k]

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...c_k 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

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

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.

思路：

这题是挑战程序设计2里的，代码按照书上写了一遍。然后仔细梳理了一下还是挺容易的。

首先建立一个结构体，里面包含p,l,r。其中p是parent，表示某节点的父结点，也就是它上面的那个。l是left，表示每个结点下最左边的那个子结点，r是right，表示每个结点右边的同行结点。

先来看输入。输入时，当第一个数字是接下来的数字的父结点，第二个数字是它有几个子结点，后面的数字就都是第一个数字的子结点了，所以从第三个数字开始，把第三个数字记作第一个数字的l，后面的记作每一个的前一个的r，并且从第三个数字开始，都要把第一个数字记作他们的p。

然后找他们的层次还是很简单的，就递归然后记一下。

最后输出。

代码：

#include<iostream>
#include<cstring>
using namespace std;
#define MAX 100005
#define NIL -1struct Node{int p,l,r;}T[MAX];
int n,D[MAX];void print(int u)
{int i,c;cout<<"node "<<u<<": ";cout<<"parent = "<<T[u].p<<", ";cout<<"depth = "<<D[u]<<", ";if(T[u].p==NIL)cout<<"root, ";else if(T[u].l==NIL)cout<<"leaf, ";else cout<<"internal node, ";cout<<"[";for(i=0,c=T[u].l;c!=NIL;i++,c=T[c].r){if(i==0)cout<<c;else cout<<", "<<c;}cout<<"]"<<endl;
}int rec(int u,int p)
{D[u]=p;if(T[u].r!=NIL)rec(T[u].r,p);if(T[u].l!=NIL)rec(T[u].l,p+1);
}int main()
{int i,j,d,v,c,l,r;while(cin>>n){for(i=0;i<n;i++)T[i].p=T[i].l=T[i].r=NIL;for(i=0;i<n;i++){cin>>v>>d;for(j=0;j<d;j++){cin>>c;if(j==0)T[v].l=c;else T[l].r=c;l=c;T[c].p=v;}}for(i=0;i<n;i++){if(T[i].p==NIL)r=i;}rec(r,0);for(i=0;i<n;i++){print(i);}}return 0;
}