Solve this interesting problem（线段树逆二分模拟的DFS递归操作）

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本文深入探讨如何解决涉及线段树的复杂问题，通过详细解释线段树的基本概念及其构建过程，指导读者掌握逆向操作模拟技巧，最终提供解决实际问题的方法。包括输入输出规范及AC代码演示。

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5323

Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1696 Accepted Submission(s): 503

Problem Description

Have you learned something about segment tree? If not, don’t worry, I will explain it for you.
Segment Tree is a kind of binary tree, it can be defined as this:
- For each node u in Segment Tree, u has two values:

Lu and

Ru .
- If

Lu=Ru , u is a leaf node.
- If

Lu≠Ru , u has two children x and y,with

Lx=Lu ,

Rx=⌊Lu+Ru2⌋ ,

Ly=⌊Lu+Ru2⌋+1 ,

Ry=Ru .
Here is an example of segment tree to do range query of sum.

Given two integers L and R, Your task is to find the minimum non-negative n satisfy that: A Segment Tree with root node's value

Lroot=0 and

Rroot=n contains a node u with

Lu=L and

Ru=R .

Input

The input consists of several test cases.
Each test case contains two integers L and R, as described above.

0≤L≤R≤109

LR−L+1≤2015

Output

For each test, output one line contains one integer. If there is no such n, just output -1.

Sample Input

Sample Output

Source

2015 Multi-University Training Contest 3

分析：要做这题，首先需要知道线段树的构建过程，然后只要对构建线段树的逆向操作进行模拟即可，DFS递归时注意左右区间边界条件。

AC code：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<map>
#include<queue>
#include<cmath>
#define LL long long
#define MAXN 1000010
using namespace std;
const LL INF=0x3f3f3f3f;
LL L,R,ans;
void dfs(LL l,LL r)
{
	if(l<=0||r>=2*R)
	{
		if(l==0)
		{
			ans=min(ans,r);
		}
		return;
	}
	int len=r-l+1;
	if(len>l)//jianzhi
		return;
	dfs(l-len,r);
	dfs(l-len-1,r);
	dfs(l,r+len);
	if(len>1)
	dfs(l,r+len-1);
}
int main()
{
	while(scanf("%I64d%I64d",&L,&R)!=EOF)
	{
		ans=INF;
		dfs(L,R);
		if(ans==INF)
			printf("-1\n");
		else
			printf("%I64d\n",ans);
	}
	return 0;
}