Catch That Cow（BFS）

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10166    Accepted Submission(s): 3179

Problem Description

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?

 

Input

Line 1: Two space-separated integers: N and K

 

Output

Line 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.

#include <cstdio>
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
int b,e;
int Mintim;
struct node
{
    int pos,t;
}k,tem;
int vis[100000+10];
queue<node> s;
void bfs()
{
    while(!s.empty())
        s.pop();
    k.pos=b,k.t=0;
    s.push(k);
    while(!s.empty())
    {
        k=s.front();
        s.pop();
        if(k.pos==e)
        {
            Mintim=k.t;
            return;
        }
        if(k.pos<0||k.pos>100000||vis[k.pos])    continue;
        vis[k.pos]=1;
        tem.t=k.t+1;
        tem.pos=k.pos+1;
        s.push(tem);
        tem.pos=k.pos-1;
        s.push(tem);
        tem.pos=k.pos*2;
        s.push(tem);
    }
}
int main()
{
    int i,j;
    freopen("in.txt","r",stdin);
    while(scanf("%d%d",&b,&e)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        bfs();
        printf("%d
",Mintim);
    }
    return 0;
}