poj3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36079		Accepted: 11123

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.

完全是BFS，没有什么技巧，主要还是用VISIT标记，防重搜

#include
#include
#include
using namespace std; 
struct tree{int x;int step;} queue[800050];
int visit[100050];
int n,k;
int bfs()
{
	int t,w,temp,minx=10000000,xx;
	t=w=1;
	queue[t].x=n;
	visit[n]=1;
	queue[t].step=0;
	while(t<=w)
	{
	
		 xx=queue[t].x;
	
		if(xx>k)
		{
		
			temp=queue[t].step+xx-k;//如果比结果大，只能减，所以可以直接求出来！
			if(temp=1&&(!visit[xx-1]))
		{
		queue[++w].x=xx-1;
		visit[xx-1]=1;
		queue[w].step=queue[t].step+1;
		if(	queue[w].x==k)
		{
			if(queue[w].step=k)
			printf("%d\n",n-k);
		else
			printf("%d\n",bfs());
	}
	
	return 0;
}