Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 30924 | Accepted: 9536 |
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 /* 功能Function Description: POJ 3278 Catch That Cow 2 开发环境Environment: DEV C++ 4.9.9.1 3 技术特点Technique: 4 版本Version: 5 作者Author: 可笑痴狂 6 日期Date: 20120730 7 备注Notes: 深搜---队列 8 */ 9 #include<iostream> 10 #include<queue> 11 #define MAX 100001 12 using namespace std; 13 14 queue<int> q; 15 bool visit[MAX]; 16 int step[MAX]; //记录步数的数组不能少 17 18 bool bound(int num) 19 { 20 if(num<0||num>100000) 21 return true; 22 return false; 23 } 24 25 int BFS(int st,int end) 26 { 27 queue<int> q; 28 int t,temp; 29 q.push(st); 30 visit[st]=true; 31 while(!q.empty()) 32 { 33 t=q.front(); 34 q.pop(); 35 for(int i=0;i<3;++i) //三个方向搜索 36 { 37 if(i==0) 38 temp=t+1; 39 else if(i==1) 40 temp=t-1; 41 else 42 temp=t*2; 43 if(bound(temp)) //越界 44 continue; 45 if(!visit[temp]) 46 { 47 step[temp]=step[t]+1; 48 if(temp==end) 49 return step[temp]; 50 visit[temp]=true; 51 q.push(temp); 52 } 53 } 54 } 55 } 56 57 int main() 58 { 59 int st,end; 60 while(scanf("%d%d",&st,&end)!=EOF) 61 { 62 memset(visit,false,sizeof(visit)); 63 if(st>=end) 64 cout<<st-end<<endl; 65 else 66 cout<<BFS(st,end)<<endl; 67 } 68 return 0; 69 }