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): 12344    Accepted Submission(s): 3833

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

题意：人抓牛，牛的位置为k，人的位置为n，人又三种走法：前进一步、后退一步、或者直接到是自己位置坐标二倍的地方，即2*n,三种走法都各花费一分钟，求人抓到牛花费的最少时间。-----BFS-----

<span style="font-family:SimSun;font-size:18px;">#include<stdio.h>#include<queue>#include<functional>#include<string.h>using namespace std;int n,k,nk,vis[100010];int d[3]={2,1,-1};  //人的三种走法 struct node{int x;int time;friend bool operator < (node a,node b) //时间小的先出队 {return a.time > b.time;}};int bfs(int t){priority_queue<node>que;node st,ed;st.x=t;st.time=0;que.push(st);while(!que.empty()){st=que.top();que.pop();if(st.x==k)return st.time;for(int i=0;i<3;i++){if(i==0)ed.x=st.x*d[i];elseed.x=st.x+d[i];if(ed.x>=100010||ed.x<0) //一定要判断 不然会出错 continue;if(!vis[ed.x]){vis[ed.x]=1;       //标记已走过的路线，以后不再走 ed.time=st.time+1;que.push(ed);}}}return 0;}int main(){while(~scanf("%d%d",&n,&k)){if(n>=k){printf("%d\n",n-k);   //要先判断，n 若大于 k 只能一步一步往回走 }else{nk=k-n;memset(vis,0,sizeof(vis));vis[n]=1;int ans=bfs(n);printf("%d\n",ans);}}}</span>