poj 3278（hdu 2717） Catch That Cow（bfs）

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?

 

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

USACO 2007 Open Silver

 

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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：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <queue>
 #define M 100005
 using namespace std;
 int n,m;
 int visit[M];    //标记数组，0表示没走过，1表示已走过
 struct node
 {
     int x,t;
 } s1,s2;
 void BFS()
 {
     queue<node> q;
     while(!q.empty())
         q.pop();
     s1.x=n;
     s1.t=;
     visit[n]=;     //农民起始位置标记为已走过
     q.push(s1);
     while(!q.empty())
     {
         s1=q.front();
         q.pop();
         if(s1.x==m)   //结束标志为农民到了牛的位置
         {
             printf("%d\n",s1.t);
             return;
         }
         s2.x=s1.x+;      //坐标+1
         if(s2.x>=&&s2.x<=M&&!visit[s2.x])  //判断变化后的数字是否超过了范围
         {
             visit[s2.x]=;
             s2.t=s1.t+;
             q.push(s2);
         }
         s2.x=s1.x-;      //坐标-1
         if(s2.x>=&&s2.x<=M&&!visit[s2.x])
         {
             visit[s2.x]=;
             s2.t=s1.t+;
             q.push(s2);
         }
         s2.x=s1.x*;      //坐标*2
         if(s2.x>=&&s2.x<=M&&!visit[s2.x])
         {
             visit[s2.x]=;
             s2.t=s1.t+;
             q.push(s2);
         }
     }
 }
 int main()
 {
     int i,j;
     while(~scanf("%d %d",&n,&m))
     {
         memset(visit,,sizeof(visit));    //开始全部定义为0
         BFS();
     }
     return ;
 }