Catch That Cow

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.

思路：BFS模板题，直接套用模板即可，看代码。

#include<stdio.h>struct note{    int x;    int s;}que[100000];int book[100000];int main(){    int i,head,tail,flag,tx,n,k;    int next[2]={1,-1};    while(scanf("%d%d",&n,&k)!=EOF)    {       for(i=0;i<100000;i++)    book[i]=0;        head=1;        tail=1;        que[tail].x=n;        que[tail].s=0;        tail++;        flag=0;         book[n]=1;        if(n!=k)        {            while(head<tail)            {                for(i=0;i<3;i++)                {                    if(i<2)                    tx=que[head].x+next[i];                    else                    tx=que[head].x*2;                    if(tx<0||tx>100000)                    continue;                    if(book[tx]==0)                    {                        book[tx]=1;                        que[tail].x=tx;                        que[tail].s=que[head].s+1;                        tail++;                                     }                    if(tx==k)                    {                        flag=1;                        break;                    }                }                if(flag==1)                break;                head++;            }            printf("%d\n",que[tail-1].s);                   }        else        printf("0\n");      }       return 0;}