HDU 2717 Catch That Cow（BFS）

Catch That Cow

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

 
      今天开始，俺终于也走上博客之路了。虽然我现在是个小白 ，但我相信 我可以在acm路上越走越好。

      话说这题好坑，最后就差一个maxn没改，竟然给我wrong answer。 我竟看了一个多小时，哎 。 

      本题为bfs，多做做bfs的题就会发现，他们都很类似，但本题需要注意，之后的next状态都是根据previous来的。

代码如下：

#include <bits/stdc++.h>
using namespace std;

#define cle(a,b) memset(a,b,sizeof(a))
#define rep(i,b) for(unsigned i=0;i<(b);i++)
int n,k;
const int MAXN=1e6;//这里是6 不是5 就是这个 坑了我好长时间</p>
bool vis[MAXN+10];

struct node
{
int x,step;
};
bool che(int x)
{
if(x<0||x>=MAXN||vis[x])
return 0;
return 1;
}
int bfs(int x)
{
queue<node> que;
node previous,next;
previous.x=x;
previous.step=0;
vis[previous.x]=1;
que.push(previous);//之后一定要push第一个状态啊 不push怎么会有呢？
while(que.size())
{
previous=que.front();/*previous在外面用 用于第一个状态加入队列 并且在while里面也用 即previous赋值了2次*/
que.pop();
if (previous.x==k)/*判断结束的条件也是previous的 previous 1次赋初值 1次赋 que.front() 1次判断 1次被next赋值 共用了4次*/
return previous.step;//最重要的  判断结束条件 return答案
next = previous;/*next只能够被previous 赋值 并且之后判断next是否能加入队列 */
next.x=previous.x+1;

if (che(next.x))//之后的便是判断语句 找符合的状态插入队列
{
next.step=previous.step+1;
vis[next.x]=1;
que.push(next);
}
next.x=previous.x-1;
if (che(next.x))//if 判断里有 1：答案的步数 应该++ 2：标记vis数组为走过 3：push进队列
{
next.step=previous.step+1;
vis[next.x]=1;
que.push(next);
}
next.x=previous.x*2;
if (che(next.x))
{
next.step=previous.step+1;
vis[next.x]=1;
que.push(next);
}
}
return -1;
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
cle(vis,0);
int ans=bfs(n);
printf("%d\n",ans);
}
return 0;
}