HDU2717:Catch That Cow(BFS)

最新推荐文章于 2024-07-06 00:20:02 发布

原创最新推荐文章于 2024-07-06 00:20:02 发布 · 226 阅读

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本文介绍了一个有趣的算法问题：一位农夫需要在数轴上通过步行或传送方式捕捉静止不动的逃逸奶牛。文章通过广度优先搜索(BFS)算法解决此问题，并提供了一段C++实现代码作为示例。

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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

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.

考虑用bfs是因为farmer的涌动整体是不规律的不具备dp的条件，即不是一个可以向上递推的过程，需要考虑所有的情况。且显然是在一种状态下由多种可行的方法，选取bfs是更加合理的。

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
const int N = 2000000;
int vis[N+10];//每个点的访问情况 
int n,k;
struct node
{
    int x,step;
};
int check(int x)
{
    if(x<0 || x>=N || vis[x])
        return 0;
    return 1;
}
int bfs(int x)
{
	queue<node>Q;
	node p,next;
	//将当前点加入判断的情况 
	p.step=0,vis[x]=1,Q.push(p);
	while(!Q.empty())
	{
		p=Q.front();
		Q.pop();
		if(p.x==k)
			return p.step; 
		//分三种情况把点加入队列等待后来的一一遍历 
		next.x=p.x+1;
		if(check(next.x))
		{
			next.step=p.step+1;
			vis[next.x]=1;
			Q.push(next);
		}
		next.x=p.x-1;
		if(check(next.x))
		{
			next.step=p.step+1;
			vis[next.x]=1;
			Q.push(next);
		}
		next.x=p.x*2;
		if(check(next.x))
		{
			next.step=p.step+1;
			vis[next.x]=1;
			Q.push(next);
		} 
	} 
	return 0;
} 
int main()
{
	int ans;
    while(~scanf("%d%d",&n,&k))
    {
        memset(vis,0,sizeof(vis));
        ans = bfs(n);
        printf("%d\n",ans);
    }
    return 0;
	return 0;
}