hdu~2717(bfs)

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7186    Accepted Submission(s): 2266

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
    
    
    
    

#include <stdio.h>
#include <string.h>
#include <queue> //用队列储存走过的点
 
using namespace std;


int sign[200005],t[200005];
int x,y;
queue<int >q;    


void bfs()
{
    while(!q.empty())
    {
        int d=q.front();
        q.pop();
        if(d==y)
            return ;

//3个判断在范围内且为走过则继续
        if(d+1>=0 && d+1<100005 && sign[d+1]==0)
        {
            q.push(d+1);
            t[d+1]=t[d]+1;
            sign[d+1]=1;
        }


        if(d-1>=0 && d-1<100005 && sign[d-1]==0)
        {
            q.push(d-1);
            t[d-1]=t[d]+1;
            sign[d-1]=1;
        }
        if(d*2>=0 && d*2<100005 && sign[d*2]==0)
        {
            q.push(d*2);
            t[2*d]=t[d]+1;
            sign[2*d]=1;
        }
    }
}
int main()
{
    while(scanf("%d %d",&x,&y)!=EOF)
    {
        while(!q.empty())
            q.pop();
        memset(sign,0,sizeof(sign)); //初始化
        memset(t,0,sizeof(t));
        q.push(x); //从第一个点开始
        sign[x]=1;
        t[x]=0; //t[]记录到达该点所学最小时间

        bfs();
        printf("%d\n",t[y]);
    }
    return 0;
}