【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[];
int D[];
void getDepth(int u,int p)//递归求结点的深度
{
D[u]=p;
if(a[u].right!=-)//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!=-)//存在最左侧子结点,深度+1
getDepth(a[u].left,p+);
}
void print(int u)
{
cout<<"node "<<u<<": parent = "<<a[u].parent<<", depth = "<<D[u]<<", ";
if(a[u].parent==-)//不存在父结点,即为根节点
cout<<"root, [";
else if(a[u].left==-)//没有子结点,即为叶
cout<<"leaf, [";
else
cout<<"internal node, [";
for(int i=,c=a[u].left; c!=-; ++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=; i<n; ++i)//初始化
a[i].left=a[i].parent=a[i].right=-;
for(int i=; i<n; ++i)
{
int id,k;
cin>>id>>k;
for(int j=; 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=; i<n; ++i)
if(a[i].parent==-)
root=i;
getDepth(root,);
for(int i=; i<n; ++i)
print(i);
}
【Aizu - ALDS1_7_A】Rooted Trees(树的表达)的更多相关文章
- 有根树的表达 Aizu - ALDS1_7_A: Rooted Trees
有根树的表达 题目:Rooted Trees Aizu - ALDS1_7_A A graph G = (V, E) is a data structure where V is a finite ...
- HDU p1294 Rooted Trees Problem 解题报告
http://www.cnblogs.com/keam37/p/3639294.html keam所有 转载请注明出处 Problem Description Give you two definit ...
- Tree - Rooted Trees
Rooted Trees A graph G = (V, E) is a data structure where V is a finite set of vertices and E is a b ...
- 10.3 Implementing pointers and objects and 10.4 Representing rooted trees
Algorithms 10.3 Implementing pointers and objects and 10.4 Representing rooted trees Allocating an ...
- TZOJ 4292 Count the Trees(树hash)
描述 A binary tree is a tree data structure in which each node has at most two child nodes, usually di ...
- HDU 1294 Rooted Trees Problem
题目大意:求有n个节点的树有几种? 题解:http://www.cnblogs.com/keam37/p/3639294.html #include <iostream> typedef ...
- Disharmony Trees 树状数组
Disharmony Trees Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Su ...
- HDU 5111 Alexandra and Two Trees 树链剖分 + 主席树
题意: 给出两棵树,每棵树的节点都有一个权值. 同一棵树上的节点的权值互不相同,不同树上节点的权值可以相同. 要求回答如下询问: \(u_1 \, v_1 \, u_2 \, v_2\):询问第一棵树 ...
- HDU1294 Rooted Trees Problem(整数划分 组合数学 DP)
讲解见http://www.cnblogs.com/IMGavin/p/5621370.html, 4 可重组合 dfs枚举子树的节点个数,相乘再累加 1 #include<iostream& ...
随机推荐
- live555 RTSP推送到Darwin出现404错误的解决
我们将Darwin部署到公网,接收live555 RTSP/RTP推送的时候,经常会出现在SETUP步骤Darwin返回404错误,经过查找原因,主要是Darwin对live555推送的sdp信息中的 ...
- bootstrap-table 行内编辑
1.文件引入 <link rel="stylesheet" href="bootstrap.css"> <link rel="sty ...
- JDBC编程步奏、问题总结(一)
jdbc编程步骤: 1. 加载数据库驱动 2. 创建并获取数据库链接 3. 创建jdbc statement对象 4. 设置sql语句 5. 设置sql语句中的参数(使用preparedStateme ...
- Mac终端基本指令,一些实用命令的收集.
基本命令1.列出文件ls 参数 目录名 例: 看看驱动目录下有什么:ls /System/Library/Extensions参数 -w 显示中文,-l 详细信息, -a 包括隐藏文件 ...
- Shell脚本学习指南笔记
Shell脚本学习指南 作者:Danbo 2015-8-3 脚本编程语言与编译型语言的差异 许多中型.大型的程序都是用编译型语言写的,例如:C.C+.Java等.这类程序只要从源代码(Source C ...
- Algorithm: Euler function
欧拉函数. phi(n)表示比n小的与n互质的数的个数,比如 phi(1) = 1; phi(2) = 1; phi(3) = 2; phi(4) = 2; phi(5) = 4; 性质: 1. 如果 ...
- 怎样使用alsa API
翻译文章的链接: http://equalarea.com/paul/alsa-audio.html 关于怎么使用ALSA API教程 这份文档帮助对ALSA API使用入门.不是一个完整的ALSA ...
- webstorm代码提示按键改为alt+/
webstorm代码提示默认按键为ctrl+空格 但是windows输入法中英文输入法的默认按键也是ctrl+空格 这就导致webstorm按键冲突,无法使用代码快捷提示按键 解决方法: 按ctrl+ ...
- html5--3.20 新增的keygen元素
html5--3.20 新增的keygen元素 学习要点 掌握fieldset/legend元素的用法(和figure和figcaption很像,只不过是作用于表单) 了解keygen元素的用法 fi ...
- swoole_http_server客户端测试
测试方法: http_server.php 文件内容 <?php // use Swoole\Http\Server; // $http = new Server("0.0.0.0&q ...