【搜索】POJ3278：Catch That Cow
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
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 <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <queue> 6 using namespace std; 7 const int maxn = 100010; 8 bool vis[maxn]; 9 int step[maxn]; 10 11 bool bound(int num) 12 { 13 if(num<0||num>100000) 14 return true; 15 return false; 16 } 17 int bfs(int sta,int endd) 18 { 19 queue<int> q; 20 int now,nextt; 21 q.push(sta); 22 vis[sta]=true; 23 while(!q.empty()) 24 { 25 now=q.front(); 26 q.pop(); 27 for(int i=0;i<3;i++) 28 { 29 if(i==0) 30 nextt=now+1; 31 else if(i==1) 32 nextt=now-1; 33 else 34 nextt=now*2; 35 if(bound(nextt)) 36 continue; 37 if(!vis[nextt]) 38 { 39 step[nextt]=step[now]+1; 40 if(nextt==endd) 41 return step[nextt]; 42 vis[nextt]=true; 43 q.push(nextt); 44 } 45 } 46 } 47 } 48 int main() 49 { 50 int sta,endd; 51 while(scanf("%d%d",&sta,&endd)!=EOF) 52 { 53 memset(vis,false,sizeof(vis)); 54 if(sta>=endd) 55 printf("%d\n",sta-endd); 56 else 57 { 58 int ans=bfs(sta,endd); 59 printf("%d\n",ans); 60 } 61 } 62 return 0; 63 }
