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); }