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

6 710 1310 11 

 

Sample Output

7-112 

 

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;}