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 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.
#include"cstdio"
#include"queue"
#include"cstring"
using namespace std;
typedef pair<int,int> P;
const int MAXN=100005;
int vis[MAXN];
int n,k;
int bfs()
{
queue<P> que;
vis[n]=1;
que.push(P(0,n));
while(!que.empty())
{
P now = que.front();que.pop();
if(now.second==k)
{
return now.first;
}
for(int i=-1;i<=1;i++)
{
int next;
if(i==0) next=now.second*2;
else next=now.second+i;
if(next>=0&&next<=MAXN&&!vis[next])//注意条件不能少，且顺序不能颠倒
{
vis[next]=1;
que.push(P(now.first+1,next));
}
}
}
return -1;
}
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
memset(vis,0,sizeof(vis));
printf("%d\n",bfs());
}
return 0;
}