poj-3278-Catch That Cow【BFS】

最新推荐文章于 2022-08-18 18:11:28 发布

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本文介绍了一种基于广度优先搜索(BFS)的算法来解决一个趣味数学问题——CatchThatCow。该问题描述了农夫约翰如何通过步行和瞬移两种方式追捕一只静止不动的逃跑奶牛，并寻求最短的时间来完成任务。文中详细解释了算法的实现过程，包括状态转移、边界判断等关键步骤。

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题目链接：点击打开链接

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 79649		Accepted: 25129

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.

思路：就是简单的 bfs 啊，有三个方向进行广搜

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
int n,k;
bool vis[100010];
struct node
{
	int x,step;
};
bool judge(int x)
{
	if(x<0||x>100000||vis[x])
		return 0;
	return 1;
}
queue<node> Q;
int bfs()
{
	node now,next;
	now.x=n;
	now.step=0;
	vis[now.x]=1;
	while(!Q.empty())	Q.pop();
	Q.push(now);
	while(!Q.empty())
	{
		now=Q.front();
		Q.pop();
		if(now.x==k)
			return now.step;
		next.x=now.x+1;
		if(judge(next.x))
		{
			next.step=now.step+1;
			vis[next.x]=1;
			Q.push(next);
		}
		next.x=now.x-1;
		if(judge(next.x))
		{
			next.step=now.step+1;
			vis[next.x]=1;
			Q.push(next);
		}
		next.x=now.x*2;
		if(judge(next.x))
		{
			next.step=now.step+1;
			vis[next.x]=1;
			Q.push(next);
		}
	}
}
int main()
{
	while(~scanf("%d%d",&n,&k))
	{
		memset(vis,0,sizeof(vis));
		printf("%d\n",bfs());
	}
	return 0;
}