[kuangbin带你飞]专题一 简单搜索-C - 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

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.

wa了两次，开始把数据域控制在0-200000之间，结果超时了，仔细考虑一下会发现，其实不会到200000，控制在0-100100就可以了

#include"iostream"
#include<queue>
#include"string.h"
using namespace std;

int n,k;
bool sign[200007];

struct node{
int x,step;
};

bool check(int a)
{
if(!sign[a]&&a>=0&&a<110000)
return true;
return false;
}

void bfs()
{
node mm,nn;
mm.x=n;
mm.step=0;
queue<node> jj;
jj.push(mm);
sign
=true;
while(!jj.empty())
{
nn=jj.front();
jj.pop();

if(nn.x==k)
{
cout<<nn.step<<endl;
return;
}

mm=nn;
mm.x++;
mm.step++;
if(check(mm.x))
{
sign[mm.x]=true;
jj.push(mm);
}

mm=nn;
mm.x--;
mm.step++;
if(check(mm.x))
{
sign[mm.x]=true;
jj.push(mm);
}

mm=nn;
mm.x=2*mm.x;
mm.step++;
if(check(mm.x))
{
sign[mm.x]=true;
jj.push(mm);
}
}
}

int main()
{

4000
cin>>n>>k;
memset(sign,0,sizeof(sign));
bfs();
return 0;
}