poj3278 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 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

这题可以用宽搜，但要注意剪枝，否则队列会爆。

#include
#include
int n,m,x2,x1;
int q[11111111],b[200005];
void bfs()
{
	int i,j,front=1,rear=1,x,xx;
	memset(q,0,sizeof(q));memset(b,0,sizeof(b));
	b[x1]=0;
	q[front]=x1;
	while(front<=rear){
		x=q[front];
		if(x==x2)break;
		front++;
			if(x+1<=100000 && b[x+1]==0){
				rear++;q[rear]=x+1;b[x+1]=b[x]+1;
			}
			if(2*x<=200000 && b[2*x]==0){
				rear++;q[rear]=2*x;b[2*x]=b[x]+1;
			}
			if(x>=1 && b[x-1]==0  ){
				rear++;q[rear]=x-1;b[x-1]=b[x]+1;
			}
			//
		}
	return;
}


int main()
{
	int i,j;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		x1=n;x2=m;
		bfs();
		printf("%d\n",b[x2]);
	}
	return 0;
}