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.

#include <iostream>
#include <vector>
#include <map>
using namespace std;
map<int, bool> mp;
int main() {
int m, n, u, v, a;
scanf("%d %d", &m, &n);
vector<int> pre(n);
for (int i = 0; i < n; i++) {
scanf("%d", &pre[i]);
mp[pre[i]] = true;
}
for (int i = 0; i < m; i++) {
scanf("%d %d", &u, &v);
for(int j = 0; j < n; j++) {
a = pre[j];
if ((a > u && a < v)|| (a > v && a < u) || (a == u) || (a == v)) break;
}
if (mp[u] == false && mp[v] == false)
printf("ERROR: %d and %d are not found.\n", u, v);
else if (mp[u] == false || mp[v] == false)
printf("ERROR: %d is not found.\n", mp[u] == false ? u : v);
else if (a == u || a == v)
printf("%d is an ancestor of %d.\n", a, a == u ? v : u);
else
printf("LCA of %d and %d is %d.\n", u, v, a);
}
return 0;
}

PAT 1143 Lowest Common Ancestor的更多相关文章

  1. 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 ...

  2. [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 ...

  3. [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 ...

  4. PAT 甲级 1143 Lowest Common Ancestor

    https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312 The lowest common ance ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. js调试技巧汇总中

    由于设备多样性(PC.ios.安卓.pad.tv)以及各设备对js脚本支持性差异性.js兼容性调试显得越来越重要. 0:尽力模仿真实场景下进行调试,迅速定位问题以及提供解决方案. 1:setTimeo ...

  2. 【转】Java实现将文件或者文件夹压缩成zip

    转自:https://www.cnblogs.com/zeng1994/p/7862288.html package com.guo.utils; import java.io.*; import j ...

  3. VC++常见错误原因解析--error LNK2019: 无法解析的外部符号 "public: void __thiscall

    根据个人遇到这个错误时的记录,原因可以分为一下几种: 原因一: 只是在.h里面声明了某个方法, 没有在cpp里面实现 . 具体讲,有时候在头文件中声明了需要的方法,确实忘记了在源文件中实现: 有时候在 ...

  4. LN : Eden Bitset_3

    Appreciation to our TA, 王毅峰, who designed this task. 问题描述 Give you N numbers a[1]...a[n] and M numbe ...

  5. 全志A33平台编译linux(分色排版)V1.1

    全志A33平台编译linux 大文实验室/大文哥 壹捌陆捌零陆捌捌陆捌贰 21504965 AT qq.com 完成时间:2017/12/13 10:41 版本:V1.1 (一)解压缩lichee备用 ...

  6. 伪装IP进行投票

    伪装IP投票说明 1,目的 在访问网页链接进行投票时,网站往往对同一个IP的投票次数进行了限制,无法连续重复投票.为此可以使用“火狐浏览器+IP修改插件”,通过人为设置浏览器IP,绕过网站IP检查,可 ...

  7. postgreSQL在Centos6下编译安装

    1.准备安装源 下载地址:https://www.postgresql.org/ftp/source/ 下载并解压. 2.软件编译安装 配置.检查安装环境 ./configure --prefix=/ ...

  8. cideogniter部署到阿里云服务器出现session加载错误

    A PHP Error was encounteredSeverity: WarningMessage: mkdir() [function.mkdir]: Invalid argumentFilen ...

  9. 7z.exe 命令行压缩文件排除文件(exclude filenames) 手记

    命令行使用格式:Usage: 7z <command> [<switches>...] <archive_name> [<file_names>...] ...

  10. Ubuntu-Python2.7安装 scipy,numpy,matplotlib

    sudo apt-get install python-scipy sudo apt-get install python-numpy sudo apt-get install python-matp ...