HDU5323 Solve this interesting problem 暴力DFS



Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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 valueLroot=0 and Rroot=n contains 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

6 710 1310 11

 

Sample Output

7-112

 

Source

2015 Multi-University Training Contest 3

题意：给定一对L,R。问是否存在一个线段树【0，n】他的一个区间为【L,R】。

思路：每次dfs，往这个区间的左边或者右边加一个区间，根据线段树的性质，左区间的长度=右区间长度，或者+1.可以将这个设置为dfs的剪枝条件。当L==0时，R就是搜到的答案之一，取最小值即可。

#include<cstdio>#include<cstring>#define LL __int64LL n;inline bool check(LL L,LL R,LL mid){    if(L<0) return 0;    if(mid!=(L+R)/2) return 0;    if(L>mid) return 0;    if(mid>=R) return 0;    return 1;}void dfs(LL L,LL R){    if(L==0)    {        if(n==-1||R<n) n=R;        return ;    }    LL tmp=R-L+1;    if(L<tmp) return;    LL Lt=L-tmp;    if(check(Lt,R,L-1)) dfs(Lt,R);    Lt--;    if(check(Lt,R,L-1)) dfs(Lt,R);    LL Rt=R+tmp;    if(check(L,Rt,R)) dfs(L,Rt);    Rt--;    if(check(L,Rt,R)) dfs(L,Rt);}int main(){    LL L,R;    while(~scanf("%I64d %I64d",&L,&R))    {        n=-1;        dfs(L,R);        printf("%I64d\n",n);    }    return 0;}