# POJ 3278 Catch That Cow (BFS)

Catch That Cow

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 82733 Accepted: 25987

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 – 1 or + 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

### 题意

1. 向前走一步（N+1）
2. 向后走一步（N-1）
3. 跳到当前位置两倍处（2*N）

### AC代码

#include <iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<queue>
#include<math.h>
#define MAXX 100005

using namespace std;

bool isvisted[MAXX];
struct node
{
int n;
int time;
node(int n,int time)
{
this->n=n;
this->time=time;
}
};
void bfs(int n,int k)
{
memset(isvisted,false,sizeof(isvisted));
node ans=node(n,0);
isvisted[n]=true;
queue<node>sk;
sk.push(ans);
while(!sk.empty())
{
node p=sk.front();
sk.pop();
isvisted[p.n]=true;     //标记走过的点
if(p.n==k)  //找到
{
ans=p;
break;
}
if(p.n+1<MAXX&&!isvisted[p.n+1])    //三种走法
sk.push(node(p.n+1,p.time+1));
if(p.n-1>=0&&!isvisted[p.n-1])
sk.push(node(p.n-1,p.time+1));
if(p.n*2<MAXX&&!isvisted[p.n*2])
sk.push(node(p.n*2,p.time+1));
}
printf("%d\n",ans.time);
}

int main()
{
int n,k;
while(~scanf("%d%d",&n,&k))
bfs(n,k);
return 0;
}