基础训练（四）BST二叉排序树

BST（poj:2309）

题目传送门
Time Limit: 1000MS
Memory Limit: 65536K

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

2
8
10

Sample Output

1 15
9 11

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cctype>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<sstream>
#include<fstream>
#include<cmath>

using namespace std;

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

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

Falling Leaves（poj:1577）

题目传送门
Time Limit: 1000MS
Memory Limit: 10000K

Description

Figure 1 shows a graphical representation of a binary tree of letters. People familiar with binary trees can skip over the definitions of a binary tree of letters, leaves of a binary tree, and a binary search tree of letters, and go right to The problem.

A binary tree of letters may be one of two things:
It may be empty.
It may have a root node. A node has a letter as data and refers to a left and a right subtree. The left and right subtrees are also binary trees of letters.

In the graphical representation of a binary tree of letters:
Empty trees are omitted completely.
Each node is indicated by
Its letter data,
A line segment down to the left to the left subtree, if the left subtree is nonempty,
A line segment down to the right to the right subtree, if the right subtree is nonempty.

A leaf in a binary tree is a node whose subtrees are both empty. In the example in Figure 1, this would be the five nodes with data B, D, H, P, and Y.

The preorder traversal of a tree of letters satisfies the defining properties:
If the tree is empty, then the preorder traversal is empty.
If the tree is not empty, then the preorder traversal consists of the following, in order
The data from the root node,
The preorder traversal of the root’s left subtree,
The preorder traversal of the root’s right subtree.

The preorder traversal of the tree in Figure 1 is KGCBDHQMPY.

A tree like the one in Figure 1 is also a binary search tree of letters. A binary search tree of letters is a binary tree of letters in which each node satisfies:

The root’s data comes later in the alphabet than all the data in the nodes in the left subtree.

The root’s data comes earlier in the alphabet than all the data in the nodes in the right subtree.

The problem:

Consider the following sequence of operations on a binary search tree of letters

Remove the leaves and list the data removed
Repeat this procedure until the tree is empty
Starting from the tree below on the left, we produce the sequence of trees shown, and then the empty tree

by removing the leaves with data

BDHPY
CM
GQ
K

Your problem is to start with such a sequence of lines of leaves from a binary search tree of letters and output the preorder traversal of the tree.

Input

The input will contain one or more data sets. Each data set is a sequence of one or more lines of capital letters.

The lines contain the leaves removed from a binary search tree in the stages described above. The letters on a line will be listed in increasing alphabetical order. Data sets are separated by a line containing only an asterisk (’*’).

The last data set is followed by a line containing only a dollar sign (’$’). There are no blanks or empty lines in the input.

Output

For each input data set, there is a unique binary search tree that would produce the sequence of leaves. The output is a line containing only the preorder traversal of that tree, with no blanks.

Sample Input

BDHPY
CM
GQ
K
*
AC
B
$

Sample Output

KGCBDHQMPY
BAC

AC代码：

#include<iostream>
#include<string>
#include<vector>

using namespace std;
char root[150];
vector<string>v;

bool check(char rt, char ch) {
	char tmp;
	while (root[rt] != 100) {
		if (rt<root[rt] && ch>root[rt])
			return false;
		else if (rt > root[rt] && ch < root[rt])
			return false;
		rt = root[rt];
	}
	return true;
}

string preorder(int levels, int leaves)
{
	if (levels < 0)
		return "";
	     //如果查询到了表的末尾，则返回空串
	int i = 0, cnt1 = 1, cnt2 = 1;
	string l, r;
	while (levels - cnt1 >= 0) {
		i = 0;
		while (i < (int)v[levels - cnt1].size() && v[levels - cnt1][i] < v[levels][leaves])i++;
		if (i > 0 && check(v[levels][leaves], v[levels - cnt1][i - 1])) {
			root[v[levels - cnt1][i - 1]] = v[levels][leaves];
			//暂时认定它为（levels，leaves）的左子树的权值
			break;
		}
		else if (i > 0 && check(v[levels][leaves], v[levels - cnt1][i - 1])) {
			root[v[levels - cnt1][i - 1]] = v[levels][leaves];
			//暂时认定它为（levels，leaves）的左子树的权值
			break;
		}
		cnt1++;
	}
	l = preorder(levels - cnt1, i - 1);
	//查找左子树
	while (levels - cnt2 >= 0) {
		i = 0;
		while (i < (int)v[levels - cnt2].size() && v[levels - cnt2][i] < v[levels][leaves])i++;
		if (i < (int)v[levels - cnt2].size() && check(v[levels][leaves], v[levels - cnt2][i])) {
			root[v[levels - cnt2][i]] = v[levels][leaves];
			          //暂时认定它为（levels，leaves）的右子树的权值  
			break;
		}
		else if (i < (int)v[levels - cnt2].size() && check(v[levels][leaves], v[levels - cnt2][i])) {
			root[v[levels - cnt2][i]] = v[levels][leaves];
			//暂时认定它为（levels，leaves）的右子树的权值  
			break;
		}
		cnt2++;
	}
	r = preorder(levels - cnt2, i);
	//查找右子树
	return v[levels][leaves] + l + r;
}

int main()
{
	string str;
	while (1) {
		v.clear();
		while (cin >> str, str != "*" && str != "$") {
			v.push_back(str);
		}
		root[v[(int)v.size() - 1][0]] = 100;
		cout << preorder((int)v.size() - 1, 0) << endl;
		if (str == "$")
			break;
	}
	return 0;
}
/*
BDHPY
CM
GQ
K
*/