搜索练习题G-07

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.

题意：

   输入两个数，用第一个数去追第二个个数，求最少步数

（第一个数只能1，变成自身两倍2，向前一步，3，向后一步）一次只能做其中一次动作。

思路：

    这就是一道很明显的广搜了，队列中依次取出，做三种动作生成三个元素放队尾，一旦出现到达第二个的值，此时必定是最少步数，细节看下面

#include<stdio.h>
 #include<queue>
 #include<string.h>
 #include<iostream>
 using namespace std;
 int a[200005];记录开始点数到此数用最少多少步，用于记忆化搜索，如曾计算过此数，便不用算了，，，这个还是看大神代码接受的指导
 int main()
 {
    int n, k, t;
    queue<int>q;
    cin>>n>>k;
    memset(a,-1,sizeof(a));
    a[n] = 0;//起始到起始当然0步
    q.push(n);
     while (!q.empty())
     {
         t=q.front();
         q.pop();
         if (t==k)
         {
             cout<<a[k]<<endl;//到达
             return 0;
         }
        if(t-1>=0&&a[t-1]==-1)//以下就是三种走法
        {
              a[t-1]=a[t]+1;q.push(t-1);
        }
        if (t+1<200001&&a[t+1]==-1)
        {
            q.push(t+1);a[t+1]=a[t]+1;
        }
        if (t*2<200004&&a[t*2]==-1)
        {
            q.push(t*2);a[t*2]=a[t]+1;
        }
    }
 }