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. tkinter的GUI设计:界面与逻辑分离(三)-- 多页面

    知识点: 使用 tkinter.Frame.tkraise() 函数去提升当前 tkinter.Frame 的 z 轴顺序,使得多个 tkinter.Frame 的可见性得以切换 本文基于:win7 ...

  2. lemon批量蒯

    RT,很久以前写的拿出来骗一骗访问量 把sh文件扔进source里面运行sh *.sh 从子目录蒯出来: #!/bin/bash for file in ./*/*/*.cpp do name=${f ...

  3. Wannafly挑战赛18B 随机数

    Wannafly挑战赛18B 随机数 设\(f_i\)表示生成\(i\)个数有奇数个1的概率. 那么显而易见的递推式:\(f_i=p(1-f_{i-1})+(1-p)f_{i-1}=(1-2p)f_{ ...

  4. AltiumDesigner 热焊盘铺铜

    在layout中,引脚与大面积的铺铜完全连接容易造成过分散热而产生虚焊以及避免因过分散热而必须使用大功率焊接器,因此在大面积铺铜时,对于接地引脚,我们经常使用热焊盘.在AltiumDesigner 中 ...

  5. Android Studio|IntelliJ IDEA 上传代码到码云

    码云 新建项目 Android Studio|IntelliJ IDEA 然后仓库就创建好了 此时左方文件应显示为红色 添加代码到git仓库 提交代码到git仓库 push等待被拒绝 拉取README ...

  6. GTK 预置对话框 GtkDialog 文件/颜色/字体选取等 GtkFileSelection

    (GTK2) 文档链接 作用:打开一个预置的对话框,如文件选取对话框 GtkFileSelection 效果下图所示 ╰── GtkDialog ├── GtkAboutDialog ├── GtkC ...

  7. Java 内存模型_2

    title: Java 内存模型_2 date: 2017-01-28 02:04:06 tags: [JMM] categories: [Programming,Java] --- Why 理解 J ...

  8. Tomcat部署与使用

    Tomcat简介 Tomcat是Apache软件基金会(Apache Software Foundation)的Jakarta 项目中的一个核心项目,由Apache.Sun和其他一些公司及个人共同开发 ...

  9. Vue 入门之组件化开发

    Vue 入门之组件化开发 组件其实就是一个拥有样式.动画.js 逻辑.HTML 结构的综合块.前端组件化确实让大的前端团队更高效的开发前端项目.而作为前端比较流行的框架之一,Vue 的组件和也做的非常 ...

  10. 慢吞吞的pip切换源

    http://blog.csdn.net/gz_liuyun/article/details/52778198