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 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
- 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 ...
- [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 ...
- [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 ...
- PAT 甲级 1143 Lowest Common Ancestor
https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312 The lowest common ance ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- SVN工具使用技巧
SVN打tag SVN打tag是一个很常用的功能,要谈打tag,还得从SVN官方推荐的目录结构说起.SVN官方推荐在一个版本库的根目录下先建立trunk.branches.tags这三个文件夹,其中t ...
- ubuntu 12.04不能mount nfs目录与挂载后只读不能写问题 (转载)
转自:http://blog.chinaunix.net/uid-20680966-id-3810455.html 服务器用的是fedora 12 以前在helper2416开发板上挂载服务器上的n ...
- bzoj 1638: [Usaco2007 Mar]Cow Traffic 奶牛交通【记忆化搜索】
震惊!记忆化搜索忘记返回map值调了半小时! 边(u,v)的经过次数是:能到u的牛数*v到n的方案数.正反两次连边,dfs两次即可 #include<iostream> #include& ...
- knockout 和mvc4结合使用
Knockout (或者Knockout.js ,KnockoutJS)是一个开源的JavaScript库,网址为www.knockoutjs.com.Knockout语法简洁.可读性好,能轻松实现与 ...
- mongodb的安装以及客户端
mongodb是一种非关系型的数据库,与传统的sql有很大的不同,但都是用于数据管理的,本人也是初学,很多地方都是模仿,在这里只是记录本人初次安装mongodb和客户端,记录一下安装的步骤,以便以后用 ...
- 四种IO模型
四种 IO 模型: 首先需要明确,IO发生在 用户进程 与 操作系统 之间.可以是客户端IO也可以是服务器端IO. 阻塞IO(blocking IO): 在linux中,默认情况下 ...
- 398 Random Pick Index 随机数索引
给定一个可能含有重复元素的整数数组,要求随机输出给定的数字的索引. 您可以假设给定的数字一定存在于数组中.注意:数组大小可能非常大. 使用太多额外空间的解决方案将不会通过测试.示例:int[] num ...
- 387 First Unique Character in a String 字符串中的第一个唯一字符
给定一个字符串,找到它的第一个不重复的字符,并返回它的索引.如果不存在,则返回 -1.案例:s = "leetcode"返回 0.s = "loveleetcode&qu ...
- Android ListView动态刷新某项Item
使用ViewHolder来刷新某项数据,而不用每次都全部刷新数据. 继承BaseAdapter,新建ViewHolder类. public class TestListAdapter extends ...
- oracle插入字符串数据时,字符串中有'单引号
使用insert into(field1,field2...) values('val1','val2'...)时,若值中有单引号时会报错. 处理方法:判断一下val1,val2中是否含有单引号,若含 ...