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题目大意：一个农民找他的牛，告诉你 他和牛的位置（牛位置保持不动），他有三种走法 往前一步，往后一步和走到他当前位置一倍的位置（卧槽这就是个bug，怎么可能跑那么快）每次动一次需要一分钟，问你最短需要多少分钟找到他的牛简单的bfs
1 #include<iostream>
2 #include<queue>
3 #include<cstring>
4 using namespace std;
5 #define M 100010
6 #define X 100000
7 struct place
8 {
9 int x,step;
10 };
11 bool vid[M];//记录是否走过
12 int main()
13 {
14 int bfs(place n,place k);
15 place n,k;
16 n.step=0;
17 while(cin>>n.x>>k.x)
18 cout<<bfs(n,k)<<endl;
19 return 0;
20 }
21 int bfs(place n,place k)
22 {
23 if(n.x==k.x)//先判断是否在相同的位置
24 return 0;
25 memset(vid,0,sizeof(vid));
26 queue<place>q;
27 place x;
28 while(1)
29 {
30 x.step=n.step+1;
31 if(n.x<k.x)//剪枝，只有当当前位置小于牛的位置才往前走
32 {
33 x.x=n.x*2;
34 if(x.x<=X&&!vid[x.x])
35 {
36 if(x.x==k.x)//bfs最好还是先判断不然会多花时间
37 return x.step;
38 q.push(x);vid[x.x]=true;
39 }
40 x.x=n.x+1;
41 if(!vid[x.x])
42 {
43 if(x.x==k.x)
44 return x.step;
45 q.push(x);vid[x.x]=1;
46 }
47 }
48 x.x=n.x-1;
49 if(!vid[x.x]&&x.x>=0)//不要以为只有大于才往后走，可能出现你往后走一步之后全翻倍走得情况
50 {
51 if(x.x==k.x)
52 return x.step;
53 q.push(x);
54 vid[x.x]=1;
55 }
56 n=q.front();
57 q.pop();
58 }
59 }
60 /*
61 附上几组比较重要的数据
62 0 100000
63 22
64 100000 0
65 100000
66 7 3072
67 10
68 */