Catch That Cow

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 88   Accepted Submission(s) : 36

Problem Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX
+ 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

Source

PKU

分析：老头要去抓住他的牛，已经知道牛的位置，和自己的位置，老头可以前进一步或者退一步，或者传送到当前位置数字的二倍的位置。计算老头需要几步能抓住他的牛。

代码：

#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;

int n,m;
int vis[200000];

struct node{
	int x,y,step;
}pp;
queue<node>q;
int  bfs()
{
	while(!q.empty())
	{
		q.pop();
	}
	q.push(pp);
	memset(vis,0,sizeof(vis));
	vis[pp.x]=1;
	while(!q.empty())
	{
		node gg=q.front();
		if(gg.x==m)
		return gg.step;
		q.pop();
		for(int i=0;i<3;i++)
		{
			node dd=gg;
			if(i==0)
			dd.x=dd.x+1;
			else if(i==1)
			dd.x=dd.x-1;
			else if(i==2)
			dd.x=dd.x*2;
			dd.step++;
			if(dd.x==m)
			return dd.step;
			if(dd.x>=0&&dd.x<=200000&&!vis[dd.x])
			{
				vis[dd.x]=1;
				q.push(dd);
			}

		}
	}
	return 0;
//	memset(bis,0,sizeof(vis));
}

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
         pp.x=n;
		 pp.step=0;
		 int time=bfs();
		 printf("%d\n",time);
	}
	return 0;
}

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