hdu 5323 Solve this interesting problem dfs 搜索

链接：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): 1470    Accepted Submission(s): 420

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=ncontains 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

6 7 10 13 10 11

 

Sample Output

7 -1 12

题意：告诉你一个线段树有 一个区间 l到r，如果有这个种线段树  问根节点  0-n，  n最小是多少。如果没有输出-1

做法：搜索剪枝，主要那个剪枝右边界比较难想 比较重要。

#include
#include
#include
#include
using namespace std;
#define inf 100000000000
typedef long long ll;
ll ans;

void dfs(ll l,ll r)
{
	if(l<0) return ;//越界
	if(l==0) {ans=min(ans,r); return;}//符合要求 取最小值
	if(l<(r-l+1)) return ;
	//l 这个区间 右边空余的长度，   r-l+1  是这个区间的长度
	//要对称，所以左边空白区间  必须是 这个区间的 某一倍数，可以相等，但是必须要大于等于 当前这个区间。
	//用于限制r的增长
	ll len=r-l+1;
	dfs(l-len,r);
	dfs(l-len-1,r);

	dfs(l,r+len);
	if(l!=r)
	dfs(l,r+len-1);
}
int main()
{
    ll l,r;
    while(scanf("%I64d%I64d",&l,&r)!=EOF){ 
		ans=inf;
		dfs(l,r);
		if(ans==inf)
			puts("-1"); 
		else
		printf("%I64d\n",ans);
	}
	return 0;
}