hdu 2717 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 pointN (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 orX + 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 andK

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.

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#include<cstdio>#include<algorithm>#include<queue>using namespace std;const int inf=0x3f3f3f3f;const int M=100010;int d[M];int n,k;int dd[2]={-1,1};int bfs(){    queue<int> que;    que.push(n);    d[n]=0;    while(que.size())    {        int a=que.front();        que.pop();        if(a==k)            break;        for(int i=0;i<3;i++)        {            if(i<2)            {                int m=a+dd[i];                if(m>=0&&m<M&&d[m]==inf)                {                    que.push(m);                    d[m]=d[a]+1;                }            }            else            {                int m=2*a;                if(m>0&&m<M&&d[m]==inf)                {                    que.push(m);                    d[m]=d[a]+1;                }            }        }    }    return d[k];}int main(){    while(~scanf("%d%d",&n,&k))    {        fill(d,d+M,inf);        printf("%d\n",bfs());    }    return 0;}