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.
广搜的基本例题：
1 #include2 #include 3 #include 4 #include 5 using namespace std; 6 struct point 7 { 8 int x;///记录位置 9 int count;///记录步数10 };11 queue q;12 struct point s,now,t;13 int vis[200000];///假设FJ开始的位置就是100000，那么变化两倍之后就是20000014 int bfs(int n,int m)15 {16 int j;17 while(!q.empty())18 {19 q.pop();20 }///清空队列21 memset(vis,0,sizeof(vis));22 vis[s.x]=1;23 q.push(s);24 while(!q.empty())25 {26 t=q.front();27 if(t.x==m)28 return t.count;29 for(j=0; j<3; j++)30 {31 now=t;32 if(j==0)33 {34 now.x=now.x+1;35 }36 else if (j==1)37 {38 now.x=now.x-1;39 }40 else if(j==2)41 {42 now.x=now.x*2;43 }44 now.count++;45 if(now.x==m)46 {47 return now.count;48 }49 if(now.x>=0&&now.x<=200000&&vis[now.x]==0)50 {51 vis[now.x]=1;52 q.push(now);53 }54 }55 q.pop();56 }57 return 0;///二者开始的位置相同58 }59 int main()60 {61 int n,m,ans;62 while(scanf("%d%d",&n,&m)!=EOF)63 {64 s.x=n;65 s.count=0;66 ans=bfs(n,m);67 printf("%d\n",ans);68 }69 return 0;70 }
反思：这道题和之前的那一道剑客救公主那一道题一样，不仅仅需要考虑题意之中的搜索方式，还要考虑搜索不到或者起始位置与终止位置相同等特殊情况，该去如何设置被调函数，该去返回一个什么样的值，这两道题都是因为这一点使我wa了好多次，引以为戒。