题目描述

由于这个跟后几周是一个系列，这周的做法比较简单。

把第一个人的所有祖先做标记，第二个人向上查找祖先直到找到一个标记过的节点或无法继续为止。

代码其实没什么意思。

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;public class Main {private static class Person {String name;String father;boolean tag = false;protected Person(String name, String father) {this.name = name;this.father = father;}}public static void main(String[] args) {Scanner in = new Scanner(System.in);int n = in.nextInt();in.nextLine();Map<String, Person> heritage = new HashMap<String, Person>();for (int i = 0; i < n; i++) {String line = in.nextLine();String[] people = line.split(" ");String father = people[0];String son = people[1];Person newPerson = new Person(son, father);heritage.remove(son);heritage.put(son, newPerson);if (heritage.get(father) == null) {Person newFather = new Person(father, null);heritage.put(father, newFather);}}int m = in.nextInt();in.nextLine();for (int i = 0; i < m; i++) {String line = in.nextLine();String[] people = line.split(" ");if (people[0].equals(people[1])) {System.out.println(people[0]);continue;}Person p1 = heritage.get(people[0]);Person p2 = heritage.get(people[1]);Person pt = p1;while (p1 != null) {pt.tag = true;if (pt.father == null) {break;} else {pt = heritage.get(pt.father);}}pt = p2;while (pt != null && pt.tag == false) {if (pt.father == null) {break;} else {pt = heritage.get(pt.father);}}if (pt == null || pt.tag == false) {System.out.println(-1);} else {System.out.println(pt.name);}pt = p1;while (p1 != null) {pt.tag = false;if (pt.father == null) {break;} else {pt = heritage.get(pt.father);}}}heritage.clear();in.close();}
}

吐槽：A和A的公共祖先是谁？A如果没出现在描述的父子关系中呢？好坑……