https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312

The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.

A binary search tree (BST) is recursively defined as a binary tree which has the following properties:

  • The left subtree of a node contains only nodes with keys less than the node's key.
  • The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
  • Both the left and right subtrees must also be binary search trees.

Given any two nodes in a BST, you are supposed to find their LCA.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 1,000), the number of pairs of nodes to be tested; and N (≤10,000), the number of keys in the BST, respectively. In the second line, N distinct integers are given as the preorder traversal sequence of the BST. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.

Output Specification:

For each given pair of U and V, print in a line LCA of U and V is A. if the LCA is found and A is the key. But if A is one of U and V, print X is an ancestor of Y.where X is A and Y is the other node. If U or V is not found in the BST, print in a line ERROR: U is not found. or ERROR: V is not found. or ERROR: U and V are not found..

Sample Input:

6 8
6 3 1 2 5 4 8 7
2 5
8 7
1 9
12 -3
0 8
99 99

Sample Output:

LCA of 2 and 5 is 3.
8 is an ancestor of 7.
ERROR: 9 is not found.
ERROR: 12 and -3 are not found.
ERROR: 0 is not found.
ERROR: 99 and 99 are not found.

代码:

#include <bits/stdc++.h>
using namespace std; int N, M;
vector<int> pre;
map<int, int> mp; int main() {
scanf("%d%d", &M, &N);
pre.resize(N);
for(int i = 0; i < N; i ++) {
scanf("%d", &pre[i]);
mp[pre[i]] = 1;
}
while(M --) {
int a, b;
scanf("%d%d", &a, &b);
if(!mp[a] && !mp[b])
printf("ERROR: %d and %d are not found.\n", a, b);
else if(!mp[a] || !mp[b])
printf("ERROR: %d is not found.\n", mp[a] ? b : a);
else {
int root;
for(int i = 0; i < N; i ++) {
root = pre[i];
if((root >= a && root <= b) || (root <= a && root >= b)) break;
} if(root == a)
printf("%d is an ancestor of %d.\n", a, b);
else if(root == b)
printf("%d is an ancestor of %d.\n", b, a);
else printf("LCA of %d and %d is %d.\n", a, b, root);
}
}
return 0;
}

  用 mp 记下是否出现过 然后只要从头找满足值在 a b 之间的就是根了

FHFHFH 过年之前最后一个工作日

PAT 甲级 1143 Lowest Common Ancestor的更多相关文章

  1. PAT甲级1143 Lowest Common Ancestor【BST】

    题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312 题意: 给定一个二叉搜索树,以及他的前 ...

  2. PAT Advanced 1143 Lowest Common Ancestor (30) [二叉查找树 LCA]

    题目 The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both ...

  3. PAT 1143 Lowest Common Ancestor[难][BST性质]

    1143 Lowest Common Ancestor(30 分) The lowest common ancestor (LCA) of two nodes U and V in a tree is ...

  4. [PAT] 1143 Lowest Common Ancestor(30 分)

    1143 Lowest Common Ancestor(30 分)The lowest common ancestor (LCA) of two nodes U and V in a tree is ...

  5. PAT 1143 Lowest Common Ancestor

    The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U ...

  6. 1143 Lowest Common Ancestor

    The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U ...

  7. 1143. Lowest Common Ancestor (30)

    The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U ...

  8. [PAT] 1143 Lowest Common Ancestor(30 分)1145 Hashing - Average Search Time(25 分)

    1145 Hashing - Average Search Time(25 分)The task of this problem is simple: insert a sequence of dis ...

  9. PAT A1143 Lowest Common Ancestor (30 分)——二叉搜索树,lca

    The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U ...

随机推荐

  1. 2017-2018-1 20155220 《信息安全系统设计基础》课下实践——实现mypwd

    学习pwd命令 输入pwd命令 于是man 1 pwd查看pwd详细 然后查看pwd实现需要的系统调用man -k; grep 在这发现了一个功能相同的内核函数getcwd 到这步就很简单了,先查看这 ...

  2. 04-cookies 会话跟踪技术

    1.会话跟踪技术 1.Http协议的无状态保存 会话理解为客户端与服务器之间的一次会晤,在一次会晤中可能会包含多次请求和响应 2 .会话路径技术使用Cookie或session完成 我们知道HTTP协 ...

  3. DELL R710使用4T硬盘亮黄灯

    事件背景 公司DELL R710的物理机上面运行的SQL SERVER数据库,因存储空间不足需要扩充空间.现系统盘(300G SAS 6Gbps 15K*2)RAID 1,数据盘(500G SAS 6 ...

  4. CF535E Tavas and Pashmaks

    今天Fakehu考的T1. 大致意思就是有n个人每个人有两个速度\(v_1,v_2\),比赛有两个路程\(A,B\),最后时间是\(A/v_1+B/v_2\).求每个人是否可能成为冠军中的一个. 显然 ...

  5. jmeter测试java代码

    有时候总是要写代码的,不得不说你也得会,这不往下看 java请求了,就的写代码,那么先来实现一个类, package com.company.jemeters; public class Hello ...

  6. Microsoft Visual Studio International Pack

    Visual Studio International Pack 包含一组类库,该类库扩展了.NET Framework对全球化软件开发的支持.使用该类库提供的类,.NET 开发人员可以更方便的创建支 ...

  7. Nginx应用场景

    1. Nginx应用场景 1)http服务器.Nginx可以独立的提供http服务,可以做网页静态服务器(也就是将静态文件放到nginx目录下,通过nginx来访问就ok)   2)虚拟主机,可以在一 ...

  8. android安卓生成密钥keystore(命令控制)

    android安卓生成密钥keystore(命令控制) • 配置JDK 详细教程 https://blog.csdn.net/u012934325/article/details/73441617/ ...

  9. python代码异常范围检查方法(非常实用)

    对于python编程的代码,如果需要进行相应的检查其中的错误或者异常,并且确定出现异常语句的大致范围,主要有以下四种方法: 1.第一种方法:遇错即止(告知原因) try  ......(所需检查语句) ...

  10. Windows Server平台 confluence6.7.1安装与破解

    1.1硬件需求建议: CPU:32/64 bit 2.27GHz双核心以上之CPU: 内存:8GB以上: 硬盘:300GB,7200转以上: 建议数据库.Confluence等各自独立一台服务器. 1 ...