POJ 2309 BST 二叉树性质位运算

最新推荐文章于 2019-07-12 16:15:00 发布

kdwycz

最新推荐文章于 2019-07-12 16:15:00 发布

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分类专栏：【OJ】数据结构【OJ】水题文章标签： POJ 2309 BST 二叉树性质位运算

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本文介绍了一种解决特定二叉搜索树问题的方法，即如何快速找到以任意节点为根的子树中的最小和最大值。通过巧妙利用二叉搜索树的性质，提出了一种简洁高效的算法，并附带了完整的C++代码实现。

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BST

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8155		Accepted: 4920

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 <= 2 ³¹ - 1).

Output

There are N lines in total, the i-th of which contains the answer for the i-th query.

Sample Input

2
8
10

Sample Output

1 15
9 11

Source

POJ Monthly,Minkerui

《数据结构编程实验》P244页。解法很巧妙。

#include <iostream>
using namespace std;

int lowbit(int x){
	return x & -x;
} 

int main()
{
	int n,x;
	cin>>n;
	while (n--){
		cin>>x;
		cout<<x-lowbit(x)+1<<' '<<x+lowbit(x)-1<<endl;
	}
	return 0;
}

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