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): 12291    Accepted Submission(s): 3816

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.
 

题意：牛和农夫在一条直线上，给定他们的位置分别为n,m。农夫有三种走法分别是往前走1步，往后走1步，瞬移到n*2的位置上。问农夫最少几步走到牛的位置上。

题解：直接bfs遍历每一种走法，走到牛的位置停止即可。注意记录步数，标记走过的位置，即数组下标大小即可。

对所有的状况进行一次搜索，最先找到的肯定就是时间最短的；

代码：

#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>using namespace std;int step[100010];int n,m;void bfs(int n,int m){memset(step,0,sizeof(step));queue<int>q;q.push(n);while(!q.empty()){int now=q.front();q.pop();if(now==m)  break;for(int i=0;i<3;i++){int next;if(i==0)  next=now+1;if(i==1)   next=now-1;if(i==2)  next=now*2;if(next>=0&&next<=100000&&!step[next]){step[next]=step[now]+1;q.push(next);}}}}int main(){while(scanf("%d%d",&n,&m)!=EOF){bfs(n,m);printf("%d\n",step[m]);}return 0;}