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
别看题简单，满满的坑啊！！你初始在n，先让你移动到k，有三种移动，—1，+1，*2。问你几步能到k。
最开始无脑搜，直接T了。后来发现bfs虽然是有点暴力的感觉但是还是要有思维在里面的！如果n>k，n只能一点点减到k。还有数组开大点，以防*2后越界。
代码如下：
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <queue> 7 using namespace std; 8 int n,k,ans; 9 bool vis[400010]; 10 struct node 11 { 12 int pos,step; 13 }; 14 bool check (int x) 15 { 16 if (vis[x]) 17 return 0; 18 if (x<0||x>400010) 19 return 0; 20 else 21 return 1; 22 } 23 void bfs (node now) 24 { 25 26 queue<node>q; 27 q.push(now); 28 while (!q.empty()) 29 { 30 node temp=q.front(); 31 q.pop(); 32 if (check(temp.pos-1)) 33 { 34 node x; 35 x.pos=temp.pos-1; 36 x.step=temp.step+1; 37 vis[x.pos]=1; 38 if (x.pos==k) 39 { 40 ans=x.step; 41 return ; 42 } 43 q.push(x); 44 } 45 if (check(temp.pos+1)&&temp.pos<k) 46 { 47 node x; 48 x.pos=temp.pos+1; 49 x.step=temp.step+1; 50 vis[x.pos]=1; 51 if (x.pos==k) 52 { 53 ans=x.step; 54 return ; 55 } 56 q.push(x); 57 } 58 if (check(temp.pos*2)&&temp.pos<k) 59 { 60 node x; 61 x.pos=temp.pos*2; 62 x.step=temp.step+1; 63 vis[x.pos]=1; 64 if (x.pos==k) 65 { 66 ans=x.step; 67 return ; 68 } 69 q.push(x); 70 } 71 } 72 } 73 int main() 74 { 75 //freopen("de.txt","r",stdin); 76 while (~scanf("%d%d",&n,&k)) 77 { 78 memset(vis,false,sizeof vis); 79 node now; 80 now.pos=n; 81 now.step=0; 82 ans=0; 83 if (n>=k) 84 printf("%d ",n-k); 85 else 86 { 87 bfs(now); 88 printf("%d ",ans); 89 } 90 } 91 return 0; 92 }