搜索-G

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.

1、题意:给你人和牛的坐标,然后人可以向前或者向后移动一步,或者向前移动到当前坐标的两倍处,让你求最少需要移动几步.
2、思路:直接广搜即可.
3、代码:

#include<iostream>#include<queue>#include<string>#include<string.h>using namespace std;int main(){int a,b,flag=0;int t=1;int tt=1;int d,dd,ddd;int ttt=1;int i;int x[200002];memset(x,0,sizeof(x));cin>>a>>b;queue<long> q;q.push(a);if(a==b)cout<<0<<endl;elsewhile(1){tt=ttt;ttt=0;for(i=0;i<tt;i++){d=q.front()+1;dd=q.front()-1;ddd=q.front()*2;if(d==b||dd==b||ddd==b){flag=1;break;}if(d>=0&&d<=200002&&x[d]==0){q.push(d);ttt++;x[d]=1;}if(dd>=0&&dd<=200002&&x[dd]==0){q.push(dd);ttt++;x[dd]=1;}if(ddd>=0&&ddd<=200002&&x[ddd]==0){q.push(ddd);ttt++;x[ddd]=1;}q.pop();}if(flag==1){cout<<t<<endl;break;}t++;}return 0;}

4、总结:一定要注意每次搜索记得将每个点标记,说明这个点走过了,不然会超时...