hdu 3278 Catch That Cow

http://poj.org/problem?id=3278

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 37107		Accepted: 11470

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

 

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
int vis[100005];
int N,K,time;
struct node
{
  int x;
  int step;
}info;
void  bfs(int n)
{
     int i;
     info.x=n;
     info.step=0;
     vis[n]=1;
     queue<node>q;
     q.push(info);
     while(!q.empty())
     {
          node pos;
          pos=q.front();
          q.pop();
          if(pos.x==K)
          {
               time=pos.step;
               return ;
          }
          for(i=0;i<3;i++)
          {
               int xx,step;
               if(i==0)
               {
                    xx=pos.x+1;
                    step=pos.step+1;
               }
               else if(i==1)
               {
                    xx=pos.x-1;
                    step=pos.step+1;

               }
               else
               {
                    xx=2*pos.x;
                    step=pos.step+1;
               }
               if(xx>=0&&xx<=100000&&!vis[xx])
               {
                    vis[xx]=1;
                    info.x=xx;
                    info.step=step;
                    q.push(info);
               }
          }
     }
}
int main()
{
     while(scanf("%d %d",&N,&K)!=EOF)
     {
          memset(vis,0,sizeof(vis));
         bfs(N);
         printf("%d\n",time);
     }
    return 0;
}