poj 3278Catch That Cow(hd2717)

最新推荐文章于 2018-07-29 16:43:26 发布

非常可乐（이녕）

最新推荐文章于 2018-07-29 16:43:26 发布

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分类专栏： DFS & BFS 文章标签：广搜

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本文介绍了一个有趣的算法问题——抓牛游戏。农民约翰要在最短时间内抓回逃跑的牛，他可以通过步行或瞬移的方式移动，每种移动方式都有特定的规则。文章通过一个具体的示例解释了如何使用广度优先搜索算法解决此问题。

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Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 61286		Accepted: 19141

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.

嗯，题目大一就是说，给n和k，n代表人所在位置，k代表cow 所在位置，人要抓住牛，但人只能一次只能走到n-1或n+1或2*n位置，每走一次花费一分钟，问最快多久能抓到牛。开始写的时候老wa后来看讨论区特殊测试数据2 2和0 10000和10000 0 最后一个不出结果，改一下才过

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int mark[100010];
int step[100010];
void bfs(int s,int e)
{
	int ss,nn;
	mark[s]=1;
	step[s]=0;
	queue<int>q;
	q.push(s);
	while(!q.empty())
	{
		ss=q.front();
		q.pop();
		if(ss==e)
		{
			printf("%d\n",step[ss]);
			return ;
		}
		nn=ss+1;
		if(nn<=100001&&!mark[nn])//就是每次这里比较的时候开始写的nn<=e
		{
			mark[nn]=1;
			step[nn]=step[ss]+1;
			q.push(nn);
		}
		nn=ss-1;
		if(nn>=0&&nn<=100001&&!mark[nn])
		{
			mark[nn]=1;
			step[nn]=step[ss]+1;
			q.push(nn);
		}
		nn=ss*2;
		if(nn<=100001&&!mark[nn])
		{
			mark[nn]=1;
			step[nn]=step[ss]+1;
			q.push(nn);
		}
	}
}
int main()
{
	int n,k;
	while(~scanf("%d%d",&n,&k))
	{
		memset(mark,0,sizeof(mark));
		memset(step,0,sizeof(step));
		bfs(n,k);
	}
	return 0;
}