Problem 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
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两个方向 三种选择 ：-1，1，2*；
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1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<cmath> 5 #include<cstdio> 6 #include<queue> 7 using namespace std; 8 int vis[100010]; 9 int n,i,j,k; 10 int cnt; 11 int main() 12 { 13 while(scanf("%d%d",&n,&k)==2) 14 { 15 queue<int>Q; 16 while(!Q.empty()) 17 Q.pop(); 18 memset(vis,0,sizeof(vis)); 19 Q.push(n); 20 while(!Q.empty()) 21 { 22 int t=Q.front(); 23 if(t==k) 24 break; 25 Q.pop(); 26 if(t>0&&!vis[t-1]) 27 { 28 int z=t-1; 29 vis[z]=vis[t]+1; 30 Q.push(z); 31 } 32 if(t<k&&!vis[t+1]) 33 { 34 int z=t+1; 35 vis[z]=vis[t]+1; 36 Q.push(z); 37 } 38 if(2*t<100005&&!vis[2*t]) 39 { 40 int z=2*t; 41 vis[z]=vis[t]+1; 42 Q.push(z); 43 } 44 45 } 46 printf("%d ",vis[k]); 47 } 48 return 0; 49 }