【Aizu - ALDS1_7_A】Rooted Trees（树的表达）

Rooted Trees

Descriptions:

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.

题目链接：

https://vjudge.net/problem/Aizu-ALDS1_7_A

题目大意：给你一个有根树的各个信息，输出它的父亲，深度，是什么性质的节点，子节点列表

输入0 - N-1节点的度和子节点(无序), 要求按照节点序号输出节点的相关信息
node id: parent = p , depth = d, type, [c1…ck] 

具体做法都在代码上

AC代码

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#define mod 1000000007
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
#define ME0(x) memset(x,0,sizeof(x))
using namespace std;
struct node
{
    int parent;
    int left,right;//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 a[100005];
int D[100005];
void getDepth(int u,int p)//递归求结点的深度
{
    D[u]=p;
    if(a[u].right!=-1)//当前结点存在右侧兄弟节点，不改变深度
        getDepth(a[u].right,p);
    if(a[u].left!=-1)//存在最左侧子结点，深度+1
        getDepth(a[u].left,p+1);
}
void print(int u)
{
    cout<<"node "<<u<<": parent = "<<a[u].parent<<", depth = "<<D[u]<<", ";
    if(a[u].parent==-1)//不存在父结点，即为根节点
        cout<<"root, [";
    else if(a[u].left==-1)//没有子结点，即为叶
        cout<<"leaf, [";
    else
        cout<<"internal node, [";
    for(int i=0,c=a[u].left; c!=-1; ++i,c=a[c].right)
    {
        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;

}
int main()
{
    int n;
    cin>>n;
    for(int i=0; i<n; ++i)//初始化
        a[i].left=a[i].parent=a[i].right=-1;
    for(int i=0; i<n; ++i)
    {
        int id,k;
        cin>>id>>k;
        for(int j=0; j<k; ++j)
        {
            int c,l;
            cin>>c;
            if(j)
                a[l].right=c;
            else
                a[id].left=c;
            l=c;
            a[c].parent=id;
        }
    }
    int root;//根节点的编号
    for(int i=0; i<n; ++i)
        if(a[i].parent==-1)
            root=i;
    getDepth(root,0);
    for(int i=0; i<n; ++i)
        print(i);
}

 

posted on 2019-06-23 22:03  Sky丨Star  阅读(394)  评论(0)  编辑  收藏  举报

导航