POJ 3278 Catch That Cow

Description

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.
题意：有一个农民和一头牛，他们在一个数轴上，牛在k位置保持不动，农户开始时在n位置。设农户当前在M位置，每次移动时有三种选择：1.移动到M-1；2.移动到M+1位置；3.移动到M*2的位置。问最少移动多少次可以移动到牛所在的位置。

#include<iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int maxn=100005;int a[maxn],time[maxn];int i,j,k,m,n;bool pd(int x){if(x<0 || x>maxn || time[x]!=0)return 0;else return 1;}int bfs(int n,int k){if(n==k)return 0;int x,f=0,l=0;a[f]=n;f++;while(f>l){x=a[l];l++;if(x+1==k || x-1==k || x*2==k)return time[x]+1;if(pd(x+1)==1){time[x+1]=time[x]+1;a[f]=x+1;f++;}if(pd(x-1)==1){time[x-1]=time[x]+1;a[f]=x-1;f++;}if(pd(x*2)==1){time[x*2]=time[x]+1;a[f]=x*2;f++;}}}int main(){cin>>n>>k;cout<<bfs(n,k);return 0;}