Catch That Cow(BFS)

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.

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
struct Point
{
  int pos,step;
}start;
queue<Point> que;
int n=0,m=0;
bool vis[200010];
int Bfs()
{
  while(!que.empty())
  {
    Point temp=que.front();
    que.pop();
    if(temp.pos==m) return temp.step;
    for(int i=1;i<=3;i++)
    {
      Point next_point=temp;
      if(i==1) next_point.pos++;
      else if(i==2) next_point.pos--;
      else if(i==3) next_point.pos*=2;
      next_point.step++;
      
      if((next_point.pos>=0&&next_point.pos<=200000)&&!vis[next_point.pos])
      {
        que.push(next_point);
        vis[next_point.pos]=1;
      }
      if(next_point.pos==m) return next_point.step;
    }
  }
}
int main()
{
  cin>>n>>m;
  start.pos = n;
  start.step=0;
  vis[start.pos]=1;
  que.push(start);
  printf("%d\n",Bfs());
  return 0;
}
点赞

发表评论

电子邮件地址不会被公开。必填项已用 * 标注

14 − 2 =