POJ--3278.Catch That Cow（BFS）

最新推荐文章于 2024-11-14 09:15:23 发布

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最新推荐文章于 2024-11-14 09:15:23 发布

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分类专栏：搜索技术文章标签： c++ 算法搜索

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本文介绍了解决 USACO 2007 Open Silver 问题中，农夫约翰如何利用步行和瞬移快速追捕逃牛的最短时间路径求解。通过广度优先搜索算法，演示了如何计算从起点到目标点所需的最少时间。

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Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 198497		Accepted: 60346

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

数组开大防止运行错误。因为int nx=dx[i]+t; d[nx]==0 这个nx的大小最大会比100,000大许多

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<queue>
using namespace std;
int d[500010];
int n,k;

void bfs()
{
	queue<int>q;
	q.push(n);
	d[n]=0;
	
	while(q.size())
	{
	   
		int t=q.front();
		q.pop();
		int dx[4]={1,-1,t};
		for(int i=0;i<3;i++)
		{
			int nx=dx[i]+t;
			
			if(d[nx]==0&&nx>=0&&nx<=100000)
			{
			//	cout<<nx<<endl;
				d[nx]=d[t]+1;
				if(nx==k)
				{
					return;
				}
				q.push(nx);
			}
		}
	}

}

int main()
{
	cin>>n>>k;
	if(k<=n)
	{
		cout<<n-k<<endl;
	}
	else
	{
		bfs();
		cout<<d[k]<<endl;
		
	}

    return 0;	
}