HDU2717-Catch That Cow-BFS

Xq_23

于 2021-01-30 15:38:45 发布

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分类专栏：搜索 #BFS 文章标签： c语言数据结构算法

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本文链接：https://blog.youkuaiyun.com/m0_46381590/article/details/113434206

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这篇博客讨论了一个数学问题，农夫约翰正试图以最快的速度捕捉到位于数轴上某个位置的逃牛。他有两种移动方式：步行和瞬移。步行时，他可以在一分钟内从任何点移动到相邻的点；瞬移时，他可以瞬间移动到当前位置的两倍。题目要求找出农夫约翰抓到逃牛所需的最短时间。例如，当农夫约翰在517的位置，而逃牛在位置17时，最快的路径是5-10-9-18-17，耗时4分钟。博客内容涉及路径规划和最优化问题。

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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.

#include<stdio.h>
#define N 100010
typedef struct 
{
	int x;
	int s;
}S;
int main()
{
	int i,n,k,h,t,f,tx;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		S d[N]={0};
		int a[N]={0};
		if(n>=k)//特判
		printf("%d\n",n-k);
		else
		{
			h=1;t=1;
	    d[t].x=n,d[t].s=0;t++;
	    a[n]=1;
	    f=0;
	    tx=n;
	    while(h<t)
	    {   
	       int e[3]={1,-1,d[h].x};//d[h].x代表当前遍历的位置
	    	for(i=0;i<3;i++)//三个方法 +1，-1，*2
	    	{
	    		tx=d[h].x+e[i];
	    		if(tx>100000)
	    		continue;
	    		if(a[tx]==0)//没有走过
	    		{
	    			a[tx]=1;//标记
					d[t].x=tx;//进队
	    			d[t].s=d[h].s+1;//+1
	    			t++;
				}
				if(tx==k)//到达k结束
				{
					f=1;
					break;
		    	}
			}
			if(f==1)
			break;
			h++;
		}
		printf("%d\n",d[t-1].s);//输出
		}
	}
	return 0;
}