hdu 2717 Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7203    Accepted Submission(s): 2274

Problem 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.
 

题意：在x坐标上，农夫在n，牛在k。农夫每次可以移动到n-1, n+1, n*2的点。求最少到达k的步数。

#include <iostream>#include <queue>using namespace std;const int Max=200010;int vis[Max];int bfs(int n,int k){    queue<int>q;    int m;    q.push(n);    vis[n]=0;    while (!q.empty())    {        n=q.front();        q.pop();        if(n==k)        {            return vis[n];        }        m=n+1;        if(m<=2*k && !vis[m])        {            q.push(m);            vis[m]=vis[n]+1;        }        m=n-1;        if(m>=0 && !vis[m])        {            q.push(m);            vis[m]=vis[n]+1;        }        m=2*n;        if(m<=2*k && !vis[m])        {            q.push(m);            vis[m]=vis[n]+1;        }    }    return -1;}int main(){    int n,k;    while (cin>>n>>k)    {        memset(vis, 0, sizeof(vis));        cout<<bfs(n, k)<<endl;    }    return 0;}