POJ 3278-Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 51790		Accepted: 16263

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.

小结：

虽然不是我做的第一道BFS题目，但是是我做过最简单的一道题目，真的要说有什么陷阱的话，感觉有2点：第一，题目明明没说，是测试多组数据（也可能是我没看清楚），但是只有按照多组测试数据来，才能AC，不然WA一辈子；第二，需要注意的就是输入1,0的这组情况，取值范围是闭区间，所以这个输出是有结果的。

恩！大概就是这样啦，发下代码就睡觉了，熬夜刷题的同志们，大家也要注意身体，早点睡觉哦！

以下是AC代码：

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<math.h>

int s,e;

int fun(int a,int x)

{

if(a==1)

return 2*x;

else if (a==2)

return x+1;

else

return x-1;

}

struct node

{

int x,step;

}point[120000];

int vis[120000];

int bfs()

{

struct node temp;

int top,end;

temp.x=s;

vis[s]=1;

temp.step=0;

top=end=0;

point[top]=temp;

while(top>=end)

{

temp=point[end];

end++;

for(int i=1;i<=3;i++)

{

struct node t;

t=temp;

t.x=fun(i,t.x);

t.step++;

if(t.x>100000||vis[t.x]||t.x<0)

continue;

if(t.x==e)

return t.step;

else

{

//printf("\n%d\n",t.x);

point[++top]=t;

vis[t.x]=1;

}

}

}

return 0;

}

int main()

{

while(~scanf("%d%d",&s,&e))

{

memset(vis,0,sizeof(vis));

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

}

return 0;

}