hdu 5323 Solve this interesting problem dfs搜索

Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1431    Accepted Submission(s): 404


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  LuRu, 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.
0LR109
LRL+12015
 

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
 

Source
2015 Multi-University Training Contest 3

给出线段树的一个区间,问满足的树【0,n】的n


由于,线段树的每个节点,的左区间可能比右区间大一,所以向上回溯的时候,就出现了四种情况,记录len为区间长度


还需要剪枝,如果r>=res 则不需要继续了,或者r>2*rr也不需要了


#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
long long l,r;
long long res;
long long MIN(long long a,long long b)
{
    if(a=res||rr>2*r) return ;
    if(ll==0)
    {
        res=rr;
        return ;
    }

    long long len=rr-ll+1;
    if(ll-len>=0)
        dfs(ll-len,rr);
    if(ll-len>=1)
        dfs(ll-len-1,rr);
    dfs(ll,rr+len);
    dfs(ll,rr+len-1);
}
int main()
{
    while(~scanf("%lld%lld",&l,&r))
    {
        res=1e16;
        dfs(l,r);
        if(res==1e16) cout<<"-1"<






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