POJ-3278 Catch the cow(BFS)

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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 andK

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<iostream>#include<cstdio>#include<queue>#include<cstring>using namespace std;const int maxx=1000010;queue<int> bfs_q;//bfs的队列 long long n,k;bool visit[2*maxx]; //设定访问数组记录该点是否访问过 int step[maxx];//记录访问到该点时的步数 int BFS(){bfs_q.push(n); //首元素压入队列 step[n]=0; visit[n]=true;while(!bfs_q.empty()){long long head,next; head=bfs_q.front(); //每次从头取数 取完就pop掉... bfs_q.pop();for(int i=-1;i<=1;i++){ //每次有三个状态 if(i==-1) next=head-1;else if(i==0) next=head+1;else next=head*2;if(next<0||next>maxx) continue; //越界处理 if(!visit[next]){ //如果该点没有被访问过 bfs_q.push(next); //把该点压入队列 visit[next]=true; //标记为已访问 step[next]=step[head]+1;}if(next==k){return step[next];}}}}int main(){scanf("%lld %lld",&n,&k);memset(visit,false,sizeof(bool));memset(step,0,sizeof(int));if(n>=k){printf("%d\n",n-k); //如果人在牛前面...直接... }else{printf("%d\n",BFS());}return 0;}