Catch That Cow+BFS

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4240    Accepted Submission(s): 1371

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.
 

 

Source

USACO 2007 Open Silver

 

Recommend

teddy

 

思路：BFS

 

#include<iostream>#include<queue>using namespace std;typedef struct infor{int x,step;}infor;int n,k,Step[100002],min1;queue<infor>q;int vi[100002];void BFS(){memset(vi,0,sizeof(vi));infor a;a.x=n;a.step=0;vi[n]=1;q.push(a);while(!q.empty()){infor b=q.front();q.pop();for(int i=0;i<3;i++){infor c=b;switch(i){case 0:{c.x+=1;c.step++;if(c.x<=100000&&!vi[c.x]){vi[c.x]=1;if(c.x==k)min1=min1>c.step?c.step:min1;q.push(c);}break;   }case 1:{c.x-=1;c.step++;if(c.x>=0&&!vi[c.x]){vi[c.x]=1;if(c.x==k)min1=min1>c.step?c.step:min1;q.push(c);}break;   }case 2:{c.x*=2;c.step++;if(c.x<=100000&&!vi[c.x]){vi[c.x]=1;if(c.x==k)min1=min1>c.step?c.step:min1;q.push(c);}break;   }}}}}int main(){while(cin>>n>>k){min1=1000000;if(n==k)cout<<"0"<<endl;else{BFS();cout<<min1<<endl;}}return 0;}