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最新推荐文章于 2018-06-01 23:18:45 发布

luckycoding

最新推荐文章于 2018-06-01 23:18:45 发布

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分类专栏： POJ 文章标签： struct output input path

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很基本的BFS题目。。。

不过开始画红线的部分没有考虑到，结果RUNTIME ERROR了，悲剧。。

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 29652		Accepted: 9148

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

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

#include"stdio.h"
int map[200002];
typedef struct
{
    int x,step;
}pos;
pos q[999999];
int main()
{
    int n,k,head=0,tail=1;
    scanf("%d%d",&n,&k);
    map[n]=1;
    q[0].x=n;
    q[0].step=0;
    while(head<tail)
    {
        pos t=q[head++];
        if(t.x==k){printf("%d\n",t.step);break;}
        if(!map[t.x+1]&&t.x<k){q[tail].step=t.step+1;q[tail++].x=t.x+1;map[t.x+1]=1;}
        if(!map[t.x-1]&&t.x>0){q[tail].step=t.step+1;q[tail++].x=t.x-1;map[t.x-1]=1;}
        if(!map[2*t.x]&&t.x<k){q[tail].step=t.step+1;q[tail++].x=2*t.x;map[2*t.x]=1;}

}
return 0;
}