【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
- EasyPusher实现Android手机屏幕桌面直播,实时推送操作画面,用于手游直播等应用
本文转自EasyDarwin开源团队成员John的博客:http://blog.csdn.net/jyt0551/article/details/52651194 由于Android 5.0提供了捕获 ...
- [数据挖掘课程笔记]人工神经网络(ANN)
人工神经网络(Artificial Neural Networks)顾名思义,是模仿人大脑神经元结构的模型.上图是一个有隐含层的人工神经网络模型.X = (x1,x2,..,xm)是ANN的输入,也就 ...
- STM32 ~ STM32 TIM重映射
复用功能 没有重映射 部分重映射 完全重映射 TIM3_CH1 PA6 PB4 PC6 CH2 PA7 PB5 PC7 CH3 PB0 PB0 PC8 CH4 PB1 PB1 PC9 /**重映射 t ...
- Xmpp学习之Asmack取经-asmack入门(一)
1.XMPPConnection:它主要是用来创建一个跟XMPP服务端的Socket连接.它是与Jabber服务端的默认连接并且已经在RFC 3920中精确定义过了.示例如下: XMPPConnect ...
- git项目.gitignore文件不生效解决办法
配置好.gitignore文件如下: HELP.md /target/ !.mvn/wrapper/maven-wrapper.jar ### STS ### .apt_generated .clas ...
- UOJ Easy Round#7
UOJ Easy Round#7 传送门:http://uoj.ac/contest/35 题解:http://matthew99.blog.uoj.ac/blog/2085 #1 题意: 在一个(2 ...
- 异常描述:hibernate懒加载中,用OpenSessionInViewFilter解决之后,同时对一个collection创建两个session访问导致异常(Illegal attempt to associate a collection with two open sessions)
在保存的时候如果使用以下方法就会报错 解决:使用merge()方法就可以解决异常... merge()方法的解释: 传入的参数在数据库中不存在的时候会添加一条数据,根据主键判断已存在的时候会更新这条数 ...
- UVA1025 A Spy in the Metro —— DP
题目链接: https://vjudge.net/problem/UVA-1025 题解: 详情请看紫书P267. 与其说是DP题,我觉得更像是模拟题,特别是用记忆化搜索写. 递推: #include ...
- 关于URL编码的一些结论
转载自:http://www.ruanyifeng.com/blog/2010/02/url_encoding.html与http://www.ruanyifeng.com/blog/2007/10/ ...
- 详解linux中install命令和cp命令的区别
基本上,在Makefile里会用到install,其他地方会用cp命令. 它们完成同样的任务——拷贝文件,它们之间的区别主要如下: .最重要的一点,如果目标文件存在,cp会先清空文件后往里写入新文件, ...