hdu5323 Solve this interesting problem(DFS关于线段树中的一些性质和规律)

Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1905    Accepted Submission(s): 573

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

 

原题

题意：一个线段树所有叶子节点编号为0~n，给出l和r，求出最小的n使得存在一个节点的左值等于l，右值等于r

关于线段树的一些性质和规律：如果存在一个节点其左值为l右值为r，那么他的父亲节点的左右值有四种可能，即

[l,2*l+1]

[l,2*r]

[2*(l-1)-r,r]

[2*(l-1)-r+1,r]

所以题目给出l和r的值后，按着四个方向进行DFS，剪枝条件：

当已找到一个合适的mi值后，r>=mi的值，return（如果只判断r>mi的话就会超时）

l<0 那肯定return了

该节点的区间长度即   r-l+1    如果大于   l      则return，  原因：将（l，r）看做这个线段树的右子树的话，左子树的长度即为l，右子树的长度为（r-l+1） 线段树性质：左子树区间长度大于等于右子树的区间长度

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
long long mi;
long long L,R;
long long minn(long long a,long long b){
return a<b?a:b;
}
void DFS(long long l,long long r){
if(l==0){
if(mi==-1) mi=r;
else mi=minn(mi,r);
}
if(mi!=-1&&r>=mi) return;
if(l<0) return;
if(r-l+1>l) return ;
DFS((l-1)*2-r,r) ;
DFS((l-1)*2+1-r,r) ;
DFS(l,r*2-l) ;
DFS(l,r*2+1-l) ;
}
int main(int argc, char *argv[])
{

while(scanf("%lld%lld",&L,&R)!=EOF){
mi=-1;
DFS(L,R);
printf("%lld\n",mi);
}
return 0;
}