hdu5323 Solve this interesting problem(爆搜)

转载请注明出处： http://www.cnblogs.com/fraud/          ——by fraud

Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1731    Accepted Submission(s): 519

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

比较隐晦的指出了爆搜的最大深度一定不会超过11层，所以，复杂度就是4^11次方，然后加一点可行性的剪枝之类，就能AC了

 //#####################
 //Author:fraud
 //Blog: http://www.cnblogs.com/fraud/
 //#####################
 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include <iostream>
 #include <sstream>
 #include <ios>
 #include <iomanip>
 #include <functional>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <list>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <set>
 #include <map>
 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <climits>
 #include <cctype>
 using namespace std;
 #define XINF INT_MAX
 #define INF 0x3FFFFFFF
 #define MP(X,Y) make_pair(X,Y)
 #define PB(X) push_back(X)
 #define REP(X,N) for(int X=0;X<N;X++)
 #define REP2(X,L,R) for(int X=L;X<=R;X++)
 #define DEP(X,R,L) for(int X=R;X>=L;X--)
 #define CLR(A,X) memset(A,X,sizeof(A))
 #define IT iterator
 typedef long long ll;
 typedef pair<int,int> PII;
 typedef vector<PII> VII;
 typedef vector<int> VI;
 ll ans = ;
 void dfs(ll l,ll r){
     if(l<)return;
     if(r<l)return;
     if(l==){
         if(ans==-)ans = r;
         else ans = min(ans,r);
         return;
     }
     if(r>=ans&&ans!=-)return;
     if((r-l+)>(l))return;
     dfs(l-(r-l)-,r);
     dfs(l,r+(r-l));
     dfs(l-(r-l)-,r);
     dfs(l,r+(r-l)+);
     return;
 }
 int main()
 {
     ios::sync_with_stdio(false);
     ll l,r;
     while(cin>>l>>r){
         ans = -;
         dfs(l,r);
         cout<<ans<<endl;
     }
     return ;
 }