Catch That Cow

最新推荐文章于 2019-04-22 11:52:24 发布

TPsilence

最新推荐文章于 2019-04-22 11:52:24 发布

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分类专栏： bfs基础搜索，队列使用文章标签： bfs

本文链接：https://blog.youkuaiyun.com/TPsilence/article/details/54411323

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本文介绍了一种解决Famer John追击固定位置逃逸奶牛问题的算法。通过结合步行与瞬间传送两种方式，利用广度优先搜索策略，找到从起点到目标点所需的最短时间。

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题目描述：

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.

解题思路：题目很容易理解，就是求一点到另一点的最小搜索次数问题。但是此题要注意界限问题和N和K的大小问题。

代码看了应该就能理解。

AC代码：

#include<stdio.h>
#include<queue>
#include<string.h>
#define s 100005
using namespace std;
int m,n;
int dr[ 2]={ 1, -1};    //前进和后退一步
int bj[s];
struct st
{
int x;   //记录当前位置
int t;   //记录搜索次数
};
int bfs(int a){
int i,dx;
   memset(bj, 0, sizeof(bj));
   queue<st> qu;
   st now,ne;
   now.x=a,now.t= 0;
   qu.push(now);
   bj[a]= 1; //标记初始搜索基点
    while(!qu.empty())
   {
   now=qu.front();
   qu.pop();
   if(now.x==n)
   return now.t; //搜索到目标点返回搜索次数
     else
   {
for(i= 0;i< 2;i++) //搜索前进一步和后退一步的点
     {
   dx=now.x+dr[i];
   if(dx>= 0&&dx<= 100000)    //不能超过题目限制
   {
   ne.x=dx;
   ne.t=now.t;
   if(dx==n)
   return now.t+ 1;
   if(bj[dx]== 0)
   {
   ne.t++;
qu.push(ne);
bj[dx]= 1;
}
}
}
dx=now.x* 2;    //搜索前进2倍的点
if(dx>= 0&&dx<= 100000)
{
ne.x=dx;
if(dx==n)
return now.t+ 1;
ne.t=now.t;
   if(bj[dx]== 0)
   {
   ne.t++;
qu.push(ne);
bj[dx]= 1;
}
}
}
}
}
int main() {
int i,j,sum,a,b;
scanf( "%d%d",&m,&n);
if(m>=n)      //此时只能后退一步
   printf( "%d\n",m-n);
else
{
sum=bfs(m);
printf( "%d\n",sum);
}
return 0;
}