Catch That Cow POJ - 3278

最新推荐文章于 2025-03-30 16:45:00 发布

原创最新推荐文章于 2025-03-30 16:45:00 发布 · 253 阅读

0 ·

CC 4.0 BY-SA版权

训练同时被 2 个专栏收录

29 篇文章

订阅专栏

bfs广搜

6 篇文章

订阅专栏

本文介绍了一道经典的算法题目，即如何通过最少步数从起点N到达终点K。允许的操作包括向左或向右移动一步及瞬间传送到当前位置的两倍位置。使用广度优先搜索算法解决该问题。

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.

题意概括：给两个数n和k，n可以+1，-1或*2，问n经过多少次运算可以得到k。

解题思路：用广搜算法，遍历每次的三种运算方法，注意m，n一开始就相同的情况。

代码：

#include<stdio.h>
#include<string.h>
struct note
{
	int x,s;
}q[250000];
int book[250000];
int main()
{
	int m,n,i,j;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		memset(book,0,sizeof(book));
		int tail=1,head=1;
		q[head].x=m;
		q[head].s=0;
		tail++;
		book[m]=1;
		int tx=m,flag=0;
		while(head<tail)
		{
			for(i=0;i<3;i++)
			{	
				if(tx==n)
				{
					flag=1;
					break;
				}
				if(i==0)
					tx=q[head].x-1;
				if(i==1)
					tx=q[head].x+1;
				if(i==2)
					tx=q[head].x*2;
				
				if(tx<0||tx>100000)
					continue;
				//printf("%d\n",tx);
				if(book[tx]==0)
				{
					book[tx]=1;
					q[tail].x=tx;
					q[tail].s=q[head].s+1;
					tail++;
				}
				
			}
			if(flag==1)
				break;
			head++;
		}
		printf("%d\n",q[tail-1].s);
	}
	return 0;
}