HDU 5323 Solve this interesting problem（DFS）

Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1507    Accepted Submission(s): 428

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

 

Source

2015 Multi-University Training Contest 3

题意：求满足输入区间的最小整数n，没有的话输出-1

思路：四种情况，直接爆搜即可

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int n,m;
int minn;

void DFS(int l,int r)
{
    if(minn && r>=minn)
    {
        return ;
    }

    if(l == 0)
    {
        if(minn)
        {
            minn = min(r,minn);
        }
        else
        {
            minn = r;
        }
        return ;
    }
    if((l+l)<(r+1))
    {
        return ;
    }
    
    DFS((l-1)*2-r,r);
    DFS((l-1)*2+1-r,r);
    DFS(l,2*r-l);
    DFS(l,2*r-l+1);

}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        minn = 0;
        if(m == 0)
        {
            printf("0\n");
            continue;
        }
        DFS(n,m);
        if(minn)
        {
            printf("%d\n",minn);
        }
        else
        {
            printf("-1\n");
        }
    }
    return 0;
}