Catch That Cow HDU

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting. 

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute 
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute. 

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

InputLine 1: Two space-separated integers: N and KOutputLine 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.Sample Input

5 17

Sample Output

4


        
 

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.
科幻大片BFS系列

#include <iostream>
using namespace std;
#include<string.h>
#include<set>
#include<stdio.h>
#include<math.h>
#include<queue>
#include<map>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
struct lll
{
    int x,bu;
}s;

int max1=210000;
int main()
{

    int n,m;
    while(cin>>n>>m)
    {
        map<int, int >a;
        queue<lll >TM;
        s.x=n;
        s.bu=0;
        a[s.x]=1;
        TM.push(s);
        while(!TM.empty())
        {
            if(TM.front().x==m)
                break;
            s.x=TM.front().x+1;
            s.bu=TM.front().bu+1;
            if(s.x>=0&&s.x<max1&&a[s.x]==0)
            {
                a[s.x]=1;
                TM.push(s);
            }
            s.x=TM.front().x-1;
            s.bu=TM.front().bu+1;
            if(s.x>=0&&s.x<max1&&a[s.x]==0)
            {
                a[s.x]=1;
                TM.push(s);
            }
            s.x=TM.front().x*2;
            s.bu=TM.front().bu+1;
            if(s.x>=0&&s.x<max1&&a[s.x]==0)
            {
                a[s.x]=1;
                TM.push(s);
            }
            TM.pop();
        }
        cout<<TM.front().bu<<endl;
    }
    return 0;
}