POJ(3278)Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 74588 Accepted: 23516

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.

一个人追一头牛，给出人的位置，与走法，问人能多少秒追到牛：

简单搜索bfs

#include <iostream>#include<algorithm>#include<stdio.h>  #include <string.h>  #include <queue>  using namespace std;  int h[1000010],v[1000010],n,k;  void bfs()  {      memset(v,0,sizeof(v));      queue<int>q;      q.push(n);      v[n]=1;      h[n]=0;      while(!q.empty())      {          int temp=q.front();          q.pop();          if(temp+1==k||temp-1==k||temp*2==k)          {              h[k]=h[temp]+1;              break;          }          if(temp-1>=0&&!v[temp-1])          {              q.push(temp-1);              v[temp-1]=1;              h[temp-1]=h[temp]+1;          }          if(temp+1<=1000000&&!v[temp+1])          {              q.push(temp+1);              v[temp+1]=1;              h[temp+1]=h[temp]+1;          }          if(temp*2<=1000000&&!v[temp*2])          {              q.push(temp*2);              v[temp*2]=1;              h[temp*2]=h[temp]+1;          }      }  }    int main()  {     while(~scanf("%d%d",&n,&k))      {          if(n==k)          {             printf("0\n");             continue;          }          else              bfs();          printf("%d\n",h[k]);     }      return 0;  }