HDU2717-Catch That Cow （BFS入门）

题目传送门：http://acm.hdu.edu.cn/showproblem.php?pid=2717

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14880    Accepted Submission(s): 4495

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.

 

题意：

从a走到b，你可以向前走一步，向后走一步或者走a*2步，求最少多少步

 

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int v[100010],t[100010];
int main()
{
    int n,a,b,i,f;
    while(scanf("%d%d",&a,&b)!=EOF)
    {
        memset(t,0,sizeof(t));
        memset(v,0,sizeof(v));
        queue<int > q;
        q.push(a);
        v[a]=1;
        if(a==b)
        {
            printf("0
");
            continue;
        }
        while(!q.empty())
        {
            int ll=q.front(),r=ll-1,l=ll+1,z=ll*2;
            q.pop();
            //printf("ll=%d
",ll);
            if(z>=0&&z<=100000&&!v[z])
            {
                q.push(z);
                v[z]=v[ll]+1;
            //    printf("v[z]=%d z=%d
",v[z],z);
            }
            if(l>=0&&l<=100000&&!v[l])
            {
                q.push(l);
                v[l]=v[ll]+1;
                //    printf("v[l]=%d l=%d
",v[l],l);
            }
            if(r>=0&&r<=100000&&!v[r])
            {
                q.push(r);
                v[r]=v[ll]+1;
                //    printf("v[r]=%d r=%d
",v[r],r);
             }
             
             if(l==b||r==b||z==b)
             {
             
                 break;
             }
                  
        }
        
        printf("%d
",v[b]-1);
        
    }
    return 0;
    
 }