河南多校暑期集训-Catch That Cow（广搜）

最新推荐文章于 2018-07-13 23:44:30 发布

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最新推荐文章于 2018-07-13 23:44:30 发布

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分类专栏：搜索文章标签：搜索

本文链接：https://blog.youkuaiyun.com/haut_ykc/article/details/52044528

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本篇探讨了FJ如何运用步行与瞬移两种方式，在最短时间内找到静止不动的逃逸奶牛。文章详细解析了一种利用队列进行广度优先搜索的算法实现，并通过样例输入输出展示了其有效性。

G - Catch That Cow

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status

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

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<limits.h>
#include<queue>
#include<algorithm>
using namespace std;
int n,m,k;
int flag[100005];
struct node
{
	int x,step;
}str,a;
int  find(int x)
{
	if(x<0 || x>100000 || flag[x])
		return 0;
	return 1;
}
int  main()
{
	int i,j;
	queue<node>q;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		memset(flag,0,sizeof(flag));
		a.step=0;
		a.x=n;
		q.push(a);
		flag[n]=1;
		while(!q.empty())
		{
			     a=q.front();
			     q.pop();
				if(a.x==k)
		        {
			         printf("%d\n",a.step);
			         break;
		        }
				str=a;
				str.step=a.step+1;
				str.x=a.x+1;
				if(find(str.x))
				{
					str.step=a.step+1;
					flag[str.x]=1;
					q.push(str);
				}
				str.x=a.x-1;
				if(find(str.x))
				{
					str.step=a.step+1;
					flag[str.x]=1;
					q.push(str);
				}
				str.x=a.x*2;
				if(find(str.x))
				{
					str.step=a.step+1;
					flag[str.x]=1;
					q.push(str);
				}
		}
	}
}