链接：https://vjudge.net/problem/POJ-3278
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.
1 #include <iostream> 2 #include <queue> 3 #include <cstring> 4 #include <cstdio> 5 using namespace std; 6 const int N = 1e5; 7 int n,k; 8 int walk[N + 10]; 9 int main() 10 { 11 scanf("%d%d",&n,&k); 12 queue<int> q; 13 q.push(n);//将起点加入队列 14 while(!q.empty()) 15 { 16 int now = q.front(); 17 q.pop(); 18 if(now == k)//找到就跳出 19 { 20 printf("%d ",walk[k]); 21 break; 22 } 23 //三种状态：+1,-1,*2 24 if(now - 1 >= 0 && !walk[now - 1]) 25 { 26 q.push(now - 1); 27 walk[now - 1] = walk[now] + 1; 28 } 29 if(now + 1 <= N && !walk[now + 1]) 30 { 31 q.push(now + 1); 32 walk[now + 1] = walk[ now ] + 1; 33 } 34 if(now*2 <= N && !walk[now*2]) 35 { 36 q.push(now*2); 37 walk[now*2] = walk[now] + 1; 38 } 39 } 40 return 0; 41 }