POJ - 3278 Catch That Cow

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

5 17

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.

代码：

#include<iostream>
#include<queue>
using namespace std;
const int MAX=1e5+5;
int dis[MAX];
int main(){
	int N,K;
	cin>>N>>K;
	queue<int>q;
	q.push(N);
	while(!q.empty()){
		int x=q.front();
		q.pop();
		if(x==K){
			cout<<dis[x]<<endl;
			break;
		}
		if(x-1>=0&&!dis[x-1]){
			q.push(x-1);
			dis[x-1]=dis[x]+1;
		}
		if(x+1<=MAX&&!dis[x+1]){
			q.push(x+1);
			dis[x+1]=dis[x]+1;
		}
		if((x<<1)<=MAX&&!dis[x<<1]){
			q.push(x<<1);
			dis[x<<1]=dis[x]+1;
		}
	}
	return 0;
}