PAT 甲级 1143 Lowest Common Ancestor
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
- 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 ...
- 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 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
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] 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 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
- spring boot 资料
http://412887952-qq-com.iteye.com/blog/2344171 http://study.163.com/course/courseMain.htm?courseId=1 ...
- c++ 方框中绘制菜单代码
绘制静态菜单 getch与getchar 接收光标控制 一.绘制静态菜单 编写函数void mainmenu( void) 二.getch与getchar getch()的作用是从键盘接收一个字 ...
- OpenStack入门篇(二十二)之实现阿里云VPC的SDN网络
1.修改/etc/neutron/neutron.conf配置 [root@linux-node1 ~]# vim /etc/neutron/neutron.conf [defalut] ... co ...
- 为什么volatile不能保证原子性?
为什么volatile能替代简单的锁,却不能保证原子性?这里面涉及volatile,是java中的一个我觉得这个词在Java规范中从未被解释清楚的神奇关键词,在Sun的JDK官方文档是这样形容vola ...
- 树莓派 Zero WH 初使用体验
12号买了一个树莓派 Zero WH,这个是什么型号呢?其实和树莓派Zero是同一系列的,加上W则表示多了无线Wifi和蓝牙模块,加上H则表示在板子上已经焊接好了2x20的排针. 这个Zero真的很迷 ...
- React——JSX
一.将表达式嵌套在JSX中 要在JSX中内嵌js表达式只需要将js表达式放在{}中,例如: const element = <h1>this is a JSX {sayName()}< ...
- Allegro16.6结构文件dxf文件的输出与导入——凡亿PCB
在pcb设计中,结构文件的导入是不可或缺的一个步骤,使用allegro16.6软件操作如下. 一.结构文件的输出 1.在allegro16.6将pcb颜色显示设置成只剩需要导出的结构. 2.看到all ...
- Revit开发-设置对象样式
最近需要读取RVT项目样板一些对象样式,翻了一下API,查找到下面的情况:
- Docker Manager for Docker Swarm deploy
一.Swarm概述 Swarm是Docker公司在2014年12月初发布的一套较为简单的工具,用来管理Docker集群,它将一群Docker宿主机变成一个单一的,虚拟的主机.Swarm使用标准的Doc ...
- [Lua] 迭代器 闭合函数 与 泛型for
首先看看一下闭合函数(closure),见如下代码: function newCounter() local i = 0 -- 非局部变量(non-local variable) return fun ...