hdoj 2717 Catch That Cow 【bfs】

Catch That Cow

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 88   Accepted Submission(s) : 36

Problem Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX+ 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

 

Source

PKU

 

分析：老头要去抓住他的牛，已经知道牛的位置，和自己的位置，老头可以前进一步或者退一步，或者传送到当前位置数字的二倍的位置。计算老头需要几步能抓住他的牛。

 

代码：

#include<cstdio>#include<algorithm>#include<queue>#include<cstring>using namespace std;int n,m;int vis[200000];struct node{int x,y,step;}pp;queue<node>q;int  bfs(){while(!q.empty()){q.pop();}q.push(pp);memset(vis,0,sizeof(vis));vis[pp.x]=1;while(!q.empty()){node gg=q.front();if(gg.x==m)return gg.step;q.pop();for(int i=0;i<3;i++){node dd=gg;if(i==0)dd.x=dd.x+1;else if(i==1)dd.x=dd.x-1;else if(i==2)dd.x=dd.x*2;dd.step++;if(dd.x==m)return dd.step;if(dd.x>=0&&dd.x<=200000&&!vis[dd.x]){vis[dd.x]=1;q.push(dd);}}}return 0;//memset(bis,0,sizeof(vis));}int main(){while(scanf("%d%d",&n,&m)!=EOF){         pp.x=n; pp.step=0; int time=bfs(); printf("%d\n",time);}return 0;}