HDOJ 5323 Solve this interesting problem BFS搜索

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Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 974    Accepted Submission(s): 263

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

6 710 1310 11

 

Sample Output

7-112

 

Source

2015 Multi-University Training Contest 3

 

/* ***********************************************Author        :CKbossCreated Time  :2015年07月28日 星期二 22时00分45秒File Name     :HDOJ5323.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;typedef long long int LL;typedef pair<LL,LL> pII;const LL INF = (1LL<<60);LL L,R;LL ans;int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);while(cin>>L>>R){ans=INF;queue<pII> q;q.push(make_pair(L,R));while(!q.empty()){pII u = q.front(); q.pop();if(u.second>ans) continue;if(u.first==0){ans=min(ans,u.second);continue;}LL deta = u.second-u.first+1;if(deta>u.first) continue;LL right = u.first-1;LL left1 = right-deta+1; LL left2 = right-deta; /// push_left1 if(left1>=0&&u.second>=0&&left1<=u.second){pII v;v.first=left1; v.second=u.second;LL D=v.second-v.first+1;if(D>v.first&&v.first) ;else q.push(v);}/// push_left2if(left2>=0&&u.second>=0&&left2<=u.second){pII v;v.first=left2; v.second=u.second;LL D=v.second-v.first+1;if(D>v.first&&v.first) ;else q.push(v);}LL left=u.second+1;LL right1=left+deta-1;LL right2=left+deta-2;/// push_right1if(u.first>=0&&right1>=0&&u.first<=right1&&right1<ans){pII v;v.first=u.first; v.second=right1;LL D=v.second-v.first+1;if(D>v.first&&v.first) ;else q.push(v);}/// push_right2if(u.first>=0&&right2>=0&&u.first<=right2&&right2<ans){pII v;v.first=u.first; v.second=right2;LL D=v.second-v.first+1;if(D>v.first&&v.first) ;else q.push(v);}}if(ans==INF) ans=-1;cout<<ans<<endl;}    return 0;}