搜索例题

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.

思想：这个用的是广度搜索。有加一，减一，乘二，三种操作。就让原来的数进行这三种操作，得到的数再进行这三种操作，一直这样下去。就好似水纹最早扩散到岸边的一定是用时最少的 ，如果再标记了那些是已验证的，就有哦能省下好多资源。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
int n,k;
int vis[100005];
int dis[100005];//
int bfs()
{
   queue<int>q;
   q.push(n);
   memset(vis,0,sizeof(vis));
   int m,t;
   vis[n]=1;
   dis[n]=0;
   while(!q.empty())
   {
      m=q.front();
      q.pop();
      if(m==k)
      {
          return dis[m];
      }
      t=m*2;
      if(t<=100000&&t>0&&!vis[t])
      {
          vis[t]=1;
          dis[t]=dis[m]+1;
          q.push(t);
      }
      t=m+1;
      if(t<=100000&&t>=0&&!vis[t])
      {
          vis[t]=1;
          dis[t]=dis[m]+1;
          q.push(t);
      }
      t=m-1;
       if(t<=100000&&t>=0&&!vis[t])
      {
          vis[t]=1;
          dis[t]=dis[m]+1;
          q.push(t);
      }


   }
   return -1;
}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        cout<<bfs()<<endl;
    }
    return 0;
}