Catch That Cow

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Catch That Cow

Time Limit:  2000MS		Memory Limit:  65536K
Total Submissions:  200198		Accepted:  60902

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

题意：n~k需要几步 ---两种走法 1：x到 x+1或x-1  2：x到2*x

Code：

#include<stdio.h>
#include<string.h>
struct node {
	int x;
	int step;
} q[222222];
int book[100100];
int main() {
	int n,k;
	while(~scanf("%d %d",&n,&k)) {
		if(n>=k)printf("%d\n",n-k);///特判 只能通过x~x-1方式到k
		else {
			memset(book,0,sizeof(book));
			int head,tail,m,flag=0;
			head=1;
			tail=1;
			q[head].x=n;
			q[head].step=0;
			tail++;///
			while(head<tail) {
				if(q[head].x==k) {
					printf("%d\n",q[head].step);
					break;
				}
				for(int i=0; i<=2; i++) {
					if(i==0)m=q[head].x+1;
					if(i==1)m=q[head].x-1;
					if(i==2)m=q[head].x*2;
					if(m>=0&&m<=100000&&!book[m]==0) {
						book[m]=1;
						q[tail].x=m;
						q[tail].step=q[head].step+1;
						tail++;
					}
				}
				head++;
			}
		}
	}
	return 0;
}