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
代码语言：javascript复制5 17Sample Output
代码语言：javascript复制4Hint
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.
这道题是关于最短步数的，首先就要想到广搜，既然想到了广搜就可以写代码了。
代码语言：javascript复制#include<iostream>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#define Swap(a,b) a^=b^=a^=b
#define cini(n) scanf("%d",&n)
#define cinl(n) scanf("%lld",&n)
#define cinc(n) scanf("%c",&n)
#define coui(n) printf("%d",n)
#define couc(n) printf("%c",n)
#define coul(n) printf("%lld",n)
#define speed ios_base::sync_with_stdio(0);//Çв»¿ÉÓÃscnaf£»
#define Max(a,b) a>b?a:b
#define Min(a,b) a<b?a:b
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=1e6 10;
const double esp=1e-9;
int m,n,x,y;
int cnt,mini=INF;
bool flag[maxn];
int bfs(int x);
int main()
{
cin>>m>>n;
cout<<bfs(m)<<endl;
}
int bfs(int x)
{
queue<pair<int,int> > t;
t.push(make_pair(x,0));
while(!t.empty())
{
pair<int,int>w=t.front();
if(w.first==n)break;
t.pop();
w.first ,w.second ;
if(w.first<0||w.first>100000||flag[w.first]);
else t.push(w),flag[w.first]=1;
w.first-=2;
if(w.first<0||w.first>100000||flag[w.first]);
else t.push(w),flag[w.first]=1;
w.first ;
w.first*=2;
if(w.first<0||w.first>100000||flag[w.first]);
else t.push(w),flag[w.first]=1;
}
return t.front().second;
}
