Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9753 Accepted Submission(s): 3054
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?
* 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.
Source
Recommend
一道很简单的一维广搜题，将每次坐标变化存入队列，标记不重复即可。
题意：一个农民去抓一头牛，输入分别为农民和牛的坐标，农民每次的移动可以坐标+1，或者坐标-1，或者坐标乘2三种变化，假设牛不知道农民来抓它而一直呆在原地不动，农民最少需要几步才能抓到牛。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 #define M 100005 6 using namespace std; 7 int n,m; 8 int visit[M]; //标记数组，0表示没走过，1表示已走过 9 struct node 10 { 11 int x,t; 12 } s1,s2; 13 void BFS() 14 { 15 queue<node> q; 16 while(!q.empty()) 17 q.pop(); 18 s1.x=n; 19 s1.t=0; 20 visit[n]=1; //农民起始位置标记为已走过 21 q.push(s1); 22 while(!q.empty()) 23 { 24 s1=q.front(); 25 q.pop(); 26 if(s1.x==m) //结束标志为农民到了牛的位置 27 { 28 printf("%d ",s1.t); 29 return; 30 } 31 s2.x=s1.x+1; //坐标+1 32 if(s2.x>=0&&s2.x<=M&&!visit[s2.x]) //判断变化后的数字是否超过了范围 33 { 34 visit[s2.x]=1; 35 s2.t=s1.t+1; 36 q.push(s2); 37 } 38 s2.x=s1.x-1; //坐标-1 39 if(s2.x>=0&&s2.x<=M&&!visit[s2.x]) 40 { 41 visit[s2.x]=1; 42 s2.t=s1.t+1; 43 q.push(s2); 44 } 45 s2.x=s1.x*2; //坐标*2 46 if(s2.x>=0&&s2.x<=M&&!visit[s2.x]) 47 { 48 visit[s2.x]=1; 49 s2.t=s1.t+1; 50 q.push(s2); 51 } 52 } 53 } 54 int main() 55 { 56 int i,j; 57 while(~scanf("%d %d",&n,&m)) 58 { 59 memset(visit,0,sizeof(visit)); //开始全部定义为0 60 BFS(); 61 } 62 return 0; 63 }