解题报告 之 HDU5323 Solve this interesting problem

解题报告 之 HDU5323 Solve this interesting 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:



and



.

- If









,
u is a leaf node.

- If









,
u has two children x and y,with









,





















,

























,









.

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













and













contains
a node u with







and







.



Input

The input consists of several test cases.

Each test case contains two integers L and R, as described above.













































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

题目大意：给出线段树中一个节点（区间[L,R]），如果原区间从0开始，问原区间右边界的最小值是多少？

分析：红果果的搜索，线段树区间的分配方法决定了向上反推的时候可能有四种情况，这样就要向四个方向搜索分别是
[l,r+len]
[l,r+len-1]
[l-len,r]
[l-len-1,r]
是因为区间长度的奇偶性造成的。
重要的剪枝是如果L已经小于0，或者已经小于length，那么则不可能是通过任何一个区间分出来的。那么停止搜索，其他的每次搜索到l==0的时候表示找到了一个新的满足要求的区间，则进行一次更新。第二个剪枝是如果r已经大于了现有的n。那么则再怎么搜索下去也不可能得到更小的答案了。

上代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;

ll n;

bool inline Check( ll l, ll r,ll mid)
{
	if(l < 0) return false;
	return true;
}

void dfs( ll l, ll r )
{
	if(l == 0 && (n == -1 || n > r))
	{
		n = r;
		return;
	}

	if(n != -1 && r >= n) return;

	ll length = r - l + 1;
	if(l < length) return;

	ll nl = l - length;
	if(Check( nl, r, l-1 ))
		dfs( nl, r );
	nl--;
	if(Check( nl, r, l-1 ))
		dfs( nl, r );

	ll nr = r + length;
	if(Check( l, nr, r ))
		dfs( l, nr );
	nr--;
	if(Check( l, nr, r ));
	dfs( l, nr );

}

int main()
{
	ll l, r;
	while(scanf( "%lld%lld", &l, &r ) == 2)
	{
		n = -1;
		dfs( l, r );
		printf( "%lld\n", n );
	}
	return 0;
}