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
此题需要注意的是queue应该定义全局，这一点上我wa了好几发
1 #include <iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<queue> 5 #include<vector> 6 #include<algorithm> 7 using namespace std; 8 struct node 9 { 10 int cur,step; 11 }st; 12 queue<node>q; 13 int m,n,ans,visit[200000]; 14 int dfs(int x) 15 { 16 int i; 17 while(!q.empty()) 18 { 19 q.pop(); 20 } 21 memset(visit,0,sizeof(visit)); 22 node s,p; 23 q.push(st); 24 visit[st.cur]=1; 25 while(!q.empty()) 26 { 27 p=q.front(); 28 q.pop(); 29 if(p.cur==n) 30 return p.step; 31 for(i=1;i<=3;i++) 32 { 33 s=p; 34 if(i==1) 35 s.cur-=1; 36 else if(i==2) 37 s.cur+=1; 38 else 39 s.cur*=2; 40 s.step++; 41 if(s.cur==n) 42 return s.step; 43 if(s.cur>=0&&s.cur<=200000&&!visit[s.cur]) 44 { 45 visit[s.cur]=1; 46 q.push(s); 47 } 48 49 } 50 } 51 52 } 53 int main() 54 { 55 while(scanf("%d%d",&m,&n)!=-1) 56 { 57 st.cur=m; 58 st.step=0; 59 ans=dfs(m); 60 printf("%d ",ans); 61 } 62 return 0; 63 }