Given a family tree for a few generations for the entire population and two people write a routine that will find out if they are blood related. Siblings are blood related since they have the same parents. Cousins are blood related since one of their parents have the same parents etc. Design the data structure first and then write the routine.

https://www.careercup.com/question?id=4812957531766784

解法:G家,we can use DFS or BFS, DFS is generally a bit easier to implement.

If they have a common ancestor.

Data structure: Graph, becausd two parents. Store parents in an array or list rather than having a separate pointer for the father and mother.

DFS/BFS startig from 1 person is not ideal. BFS/Iterative Deepening for common ancestor from each person, as don't process unnecessary children.

Lots of ways to determine a common ancestor. pure recursion, recursively compute the ancestor set of both people and check intersection, iterative deepening to build up the sets.

Java:

For 2 persons to be blood related, perform a Breadth First Traversal with the person as root and the parents as child nodes and store in a ArrayList.
Then for both the persons search is there is a common parent in the ArrayList.
If common parent found then, the persons are related, else not related.

The data structure to store the generation tree will have nodes defined as

public class Person {
String name;
ArrayList<Person> children = null;
Person parent1;
Person parent2; public Person(String personName, Person personParent1, Person personParent2) {
name = personName;
children = new ArrayList<Person>();
parent1 = personParent1;
parent2 = personParent2;
}
public Person getParent() { return parent; }
}

Java:

public class Person {
Person[] parents;
} // naming for cousins is: n th cousin m times removed
// where n is the min generations to a common ancestor and m is the number of generations difference between the 2 cousins
// so this is going to be O((2^n+m)+2) which is still more efficient than dfs assuming the num generations in the population is > n+m
public boolean bloodRelated(Person p1, Person p2) {
// simple search would go down p1's children/grandchildren/etc and see if we find p2
// then vice versa
// then worry about cousin style relationships
// here we'd go up the parent tree on both until we found a common node (or ran out of data) // we could take this last approach anyway and it would get us a parent-child match too
Set<Person> p1Ancestors = new HashSet<Person>();
Set<Person> p2Ancestors = new HashSet<Person>(); // so ideally here we're going to do BFS, but we're going to do 2 at once to try to minimise the depth we have to go
List<Person> p1Discovered = new LinkedList<Person>();
p1Discovered.add(p1);
List<Person> p2Discovered = new LinkedList<Person>();
p2Discovered.add(p2); while (!p1Discovered.isEmpty() || !p2Discovered.isEmpty()) {
Person nextP1 = p1Discovered.remove(0);
if (nextP1 != null) {
if (p2Ancestors.contains(nextP1)) {
return true;
} for (Person parent : nextP1.parents) {
p1Discovered.add(parent);
}
p1Ancestors.add(nextP1);
} Person nextP2 = p2Discovered.remove(0);
if (nextP2 != null) {
if (p1Ancestors.contains(nextP2)) {
return true;
} for (Person parent : nextP2.parents) {
p2Discovered.add(parent);
}
p2Ancestors.add(nextP2);
}
}
return false;
}

  

Given a family tree, find out if two people are blood related的更多相关文章

  1. openerp学习笔记 context 的应用

    1.在Action中定义,context用于传递搜索条件和分组条件,在搜索视图中默认显示: 示例代码: <record model="ir.actions.act_window&quo ...

  2. 日常训练 dfs 之 拓扑排序

    今天被拓扑排序给折磨了一天,主要就是我的一个代码有点小bug,真难找... 先来看看我今天写的题目吧! C. Fox And Names Fox Ciel is going to publish a ...

  3. python 自然语言处理(四)____词典资源

    词典或者词典资源是一个词和/或短语及其相关信息的集合,例如:词性和词意定义等相关信息.词典资源附属于文本,而且通常在文本的基础上创建和丰富.下面列举几种nltk中的词典资源. 1. 词汇列表语料库 n ...

  4. [数据结构]——二叉树(Binary Tree)、二叉搜索树(Binary Search Tree)及其衍生算法

    二叉树(Binary Tree)是最简单的树形数据结构,然而却十分精妙.其衍生出各种算法,以致于占据了数据结构的半壁江山.STL中大名顶顶的关联容器--集合(set).映射(map)便是使用二叉树实现 ...

  5. SAP CRM 树视图(TREE VIEW)

    树视图可以用于表示数据的层次. 例如:SAP CRM中的组织结构数据可以表示为树视图. 在SAP CRM Web UI的术语当中,没有像表视图(table view)或者表单视图(form view) ...

  6. 无限分级和tree结构数据增删改【提供Demo下载】

    无限分级 很多时候我们不确定等级关系的层级,这个时候就需要用到无限分级了. 说到无限分级,又要扯到递归调用了.(据说频繁递归是很耗性能的),在此我们需要先设计好表机构,用来存储无限分级的数据.当然,以 ...

  7. 2000条你应知的WPF小姿势 基础篇<45-50 Visual Tree&Logic Tree 附带两个小工具>

    在正文开始之前需要介绍一个人:Sean Sexton. 来自明尼苏达双城的软件工程师.最为出色的是他维护了两个博客:2,000Things You Should Know About C# 和 2,0 ...

  8. Leetcode 笔记 110 - Balanced Binary Tree

    题目链接:Balanced Binary Tree | LeetCode OJ Given a binary tree, determine if it is height-balanced. For ...

  9. Leetcode 笔记 100 - Same Tree

    题目链接:Same Tree | LeetCode OJ Given two binary trees, write a function to check if they are equal or ...

随机推荐

  1. Django 第一天 开端

    今日内容: 一.HTTP协议 1.HTTP协议简介 参考博客:https://www.cnblogs.com/clschao/articles/9230431.html 是超文本传输协议 现在使用最广 ...

  2. NIO原理详解

    版权声明:本文为博主原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接和本声明.本文链接:https://blog.csdn.net/CharJay_Lin/article/d ...

  3. ClickHouse 分布式高可用集群搭建(转载)

    一.ClickHouse安装方式: 源码编译安装 Docker安装 RPM包安装 为了方便使用,一般采用RPM包方式安装,其他两种方式这里不做说明. 二.下载安装包 官方没有提供rpm包,但是Alti ...

  4. Linux rpm安装指定安装路径

    可以使用prefix参数. rpm -i –prefix=/home/gpadmin    greenplum-db-6.0.0-rhel6-x86_64.rpm 将greenplum-db-6.0. ...

  5. HDP 企业级大数据平台

    一 前言 阅读本文前需要掌握的知识: Linux基本原理和命令 Hadoop生态系统(包括HDFS,Spark的原理和安装命令) 由于Hadoop生态系统组件众多,导致大数据平台多节点的部署,监控极其 ...

  6. Linux 内存Cache和Buffer理解

    在 Linux 系统中,我们经常用 free 命令来查看系统内存的使用状态.在一个 RHEL6 的系统上,free 命令的显示内容大概是这样一个状态:   [root@tencent64 ~]# fr ...

  7. 多个List 或 Array 进行 合并

    1.调用 var aaa = new List<string>() { "0" }; var a1 = new List<string>() { " ...

  8. Linux 文件与目录的权限

    文件默认权限:umask umask就是指定“目前用户在新建文件或目录时候的权限默认值”.查询umask有以下两种方式: 目录与文件的默认权限是不一样的: - 文件,默认没有可执行(x)权限,只有r. ...

  9. 11.linux dns服务器建立和安装apache

    dns服务器建立 1.安装bind建立dns服务器     yum install bind -y 2.安装好修改配置文件:vim /etc/named.conf 修改:      listen-on ...

  10. distribution system index

    Resiliency:可译为容错性,强调从错误状态恢复的能力.形容词Resilient可译作“可容错的”. Elasticity:可译为伸缩性,强调根据负载进行水平伸缩的能力.形容词Elastic可译 ...