POJ  3278  Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24664		Accepted: 7604

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

泪奔了，跑了797MS，简单BFS，注意n==k的情况，还有如果n>=k，直接就可以判断是n-k了

代码（附测试数据）：

#include<stdio.h>

typedef struct point{

int x,step;

}target;

target que[5000000];

int map[5000000]={0},k;

int BFS(int n)

{

int i,end,top;

target next,in;

if(n==k)

return 0;

end=top=0;

que[top].x=n;

que[top].step=0;

while(top>=end)

{

in=que[end];

end=(end+1)% 5000000;

for(i=0;i<3;i++)

{

if(i==0)

next.x=in.x+1;

else if(i==2)

next.x=in.x-1;

else next.x=in.x*2;

if(next.x==k)

return in.step+1;

if(next.x>=0&&next.x<5000000&&map[next.x]==0)

{

map[next.x]=1;

top=(top+1)% 5000000;

next.step=in.step+1;

que[top]=next;

}

}

}

return 0;

}

int main()

{

int num,n;

scanf("%d%d",&n,&k);

map[n]=1;

if(k<=n)

num=n-k;

else

num=BFS(n);

printf("%d\n",num);

return 0;

}

测试数据：

1 1

0

1 5

3

5 1

4

1 100000

21

200 100000

16

13 15000

15

21 4500

14

64 7913

12

15 438

8

159 753

32

486 4267

52

147 963

31

486 15347

14