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.

問題連接：http://poj.org/problem?id=3278

問題分析：利用BFS，一層一層搜索，知道找到牛的位置

AC代碼：

#include<iostream>#include<cstring>#include<queue>#define p 100010using namespace std;int visit[p];int n,k;struct Catch{int step;int x;Catch(int xx,int s):x(xx),step(s){}};queue<Catch> q;int main(){scanf("%d %d",&n,&k);memset(visit,0,sizeof(visit));q.push(Catch(n,0));visit[n]=1;while(!q.empty()){Catch c=q.front();q.pop();if(c.x==k){printf("%d\n",c.step);return 0;}else{if(c.x-1>=0&&!visit[c.x-1]){q.push(Catch(c.x-1,c.step+1));visit[c.x-1]=1;//printf("c1:%d\n",c.step+1);}if(c.x+1<p&&!visit[c.x+1]){q.push(Catch(c.x+1,c.step+1));visit[c.x+1]=1;//printf("c2:%d\n",c.step+1);}if(c.x*2<p&&!visit[c.x*2]){q.push(Catch(c.x*2,c.step+1));visit[c.x*2]=1;//printf("c3:%d\n",c.step+1);}}}}

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