文章标题 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.

从N到M最少几步；

两种走法：

(1)当前数加1或者减1；

(2)当前数*2；

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
int N,M;
int b[200009];
struct node
{
int x,ans;
};
int bfs(int x,int ans)
{
node xy,xz;
xy.x=x,xy.ans=ans;
queue<node>s;
s.push(xy);
while(!s.empty())
{
xy=s.front();
s.pop();
if(xy.x==M)
{
return xy.ans;
}
for(int i=1;i<=3;i++)
{
if(i==1)
{
int xi=xy.x-1;
if(!b[xi]&&xi>=0&&xi<=100000)
{
b[xi]=1;
xz.x=xi,xz.ans=xy.ans+1;
s.push(xz);
}
}
if(i==2)
{
int xi=xy.x+1;
if(!b[xi]&&xi>=0&&xi<=100000)
{
b[xi]=1;
xz.x=xi,xz.ans=xy.ans+1;
s.push(xz);
}
}
if(i==3)
{
int xi=xy.x*2;
if(!b[xi]&&xi>=0&&xi<=100000)
{
b[xi]=1;
xz.x=xi,xz.ans=xy.ans+1;
s.push(xz);
}
}
}
}
return 0;
}
int main()
{
while(~scanf("%d%d",&N,&M))
{
memset(b,0,sizeof(b));
int step=bfs(N,0);
printf("%d\n",step);
}
return 0;
}