首页 > 代码库 > POJ 2309 BST(二叉搜索树)
POJ 2309 BST(二叉搜索树)
BST
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 8657 | Accepted: 5277 |
Description
Consider an infinite full binary search tree (see the figure below), the numbers in the nodes are 1, 2, 3, .... In a subtree whose root node is X, we can get the minimum number in this subtree by repeating going down the left node until the last level, and we can also find the maximum number by going down the right node. Now you are given some queries as "What are the minimum and maximum numbers in the subtree whose root node is X?" Please try to find answers for there queries.
Input
In the input, the first line contains an integer N, which represents the number of queries. In the next N lines, each contains a number representing a subtree with root number X (1 <= X <= 231 - 1).
Output
There are N lines in total, the i-th of which contains the answer for the i-th query.
Sample Input
2810
Sample Output
1 159 11
算法分析:
此题目的意思是,按照原则建成的二叉树,每次我输入一个根节点的值,从当前根节点一直往左跑,跑到左边最底层,此时的该点,一定是以输入值为根节点的二叉树的
最小值,同理往右边跑,一直跑到底,则是最大值。此题目符合树状数组的结构特点。
#include <iostream>#include <stdio.h>#include <string>#include <string.h>#include <iomanip>#include <algorithm>using namespace std;int main(){ int t; int n;int m, dd; scanf("%d", &t); while(t--) { scanf("%d", &m); dd= m&(-m) ; dd--; //printf("%d\n", dd); printf("%d %d\n", m-dd, m+dd ); } return 0;}
POJ 2309 BST(二叉搜索树)
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。