hdu-2717Catch That Cow（bfs 求最少几步达到指定值）

最新推荐文章于 2020-02-26 16:54:30 发布

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本文介绍了一个有趣的算法问题——如何让农夫约翰用最短的时间抓到一只静止不动的逃逸奶牛。通过步行和瞬移两种方式，在一条数轴上从起点N到达奶牛所在位置K。文中提供了一种使用广度优先搜索解决此问题的方法。

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Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12531 Accepted Submission(s): 3876

Problem 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.

Source

USACO 2007 Open Silver

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#include<iostream>
#include<string>
#include<cstring>
#include<queue>
#include<cstdio>
#include<cmath>
using namespace std;
int vis[200100];
int n,m;
struct node{
	int first;
	int second;
	node(int a,int b)
	{
		first=a;
		second=b;
	}
};
void bfs()
{
	queue<node> que;
	que.push(node (n,0));
	vis[n]=1;
	int k,t;
	while(1)
	{
		node now=que.front();
		que.pop();
		k=now.first;
		t=now.second;
		if(k==m)
		{
			printf("%d\n",t);
			break;
		}
		if(!vis[k-1]&&k-1>=0)
		{
			vis[k-1]=1;
			que.push(node(k-1,t+1));
		}
		if(!vis[k+1]&&k+1<=100000)
		{
			vis[k+1]=1;
			que.push(node(k+1,t+1));
		}
		if(!vis[k*2]&&k*2<=100000)
		{
			vis[k*2]=1;
			que.push(node(k*2,t+1));
		}
	}
}
int main()
{
	while(~scanf("%d%d",&n,&m))
	{
		memset(vis,0,sizeof(vis));
		bfs();
	}
	return 0;
}