POJ 3278 Catch That Cow（宽搜）

最新推荐文章于 2024-01-18 14:10:50 发布

原创最新推荐文章于 2024-01-18 14:10:50 发布 · 165 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#宽搜

搜索专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一个经典的最短路径问题——POJ3278追捕奶牛。农夫约翰通过步行或瞬移的方式在数轴上追捕不动的奶牛，使用广度优先搜索算法寻找从起始位置到目标位置的最短步数。

POJ 3278 Catch That Cow
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.

题意：农夫约翰要追他的牛，开始给你约翰的位置和牛的位置，约翰一次可以向前移一步，或者向后移一步，或者移动到当前坐标点*2的位置，问最少移动多少步到达牛的位置。

题解：
简单bfs，但是需要开个vis数组记录哪些点已经到过，防止塞进队列的数过多。

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
struct cc{
	int x,step;
};
queue<cc>q;
bool vis[2000000];
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	q.push((cc){n,0});
	int ans=0;
	vis[n]=1;
	while(!q.empty())
	{
		cc u=q.front(); q.pop();
		if(u.x==m)
		{
			ans=u.step;
			break;
		}
		if(u.x-1>=0&&!vis[u.x-1])
		{
			q.push((cc){u.x-1,u.step+1});
			vis[u.x-1]=1;
		}
		if(u.x+1<=100000&&!vis[u.x+1])
		{
			q.push((cc){u.x+1,u.step+1});
			vis[u.x+1]=1;
		}
		if(u.x*2<=100000&&!vis[u.x*2])
		{
			q.push((cc){u.x*2,u.step+1});
			vis[u.x*2]=1;
		}
	}
	printf("%d",ans);
	return 0;
}