抓牛问题
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Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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>#define N 100005int u,v,x,f[N],q[N*2],h,t;int main(){scanf("%d%d",&u,&v);q[t++]=u;while(h<t){u=q[h++];if(u==v){printf("%d\n",f[u]);break;}x=f[u]+1;if(u>0&&!f[u-1])f[q[t++]=u-1]=x;if(u<=N&&!f[u+1])f[q[t++]=u+1]=x;if(u*2<=N&&!f[u*2])f[q[t++]=u*2]=x;}}
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