Catch That Cow POJ

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.

题意概括：给两个数n和k，n可以+1，-1或*2，问n经过多少次运算可以得到k。

解题思路：用广搜算法，遍历每次的三种运算方法，注意m，n一开始就相同的情况。

代码：

#include<stdio.h>#include<string.h>struct note{int x,s;}q[250000];int book[250000];int main(){int m,n,i,j;while(scanf("%d%d",&m,&n)!=EOF){memset(book,0,sizeof(book));int tail=1,head=1;q[head].x=m;q[head].s=0;tail++;book[m]=1;int tx=m,flag=0;while(head<tail){for(i=0;i<3;i++){if(tx==n){flag=1;break;}if(i==0)tx=q[head].x-1;if(i==1)tx=q[head].x+1;if(i==2)tx=q[head].x*2;if(tx<0||tx>100000)continue;//printf("%d\n",tx);if(book[tx]==0){book[tx]=1;q[tail].x=tx;q[tail].s=q[head].s+1;tail++;}}if(flag==1)break;head++;}printf("%d\n",q[tail-1].s);}return 0;}