Catch That Cow

                                            Catch That Cow                            Time Limit: 2000MS      Memory Limit: 65536K                            Total Submissions: 40350        Accepted: 12560

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.

最典型的搜索，用了一个结构体存步数和位置，用队列比较好理解，从n开始向下一层一层的遍历，进队，判断，出队。别忘了判重

#include<iostream>#include<queue>#include<algorithm>#include<string.h>using namespace std;int vis[100010];struct x{    int loc;    int step;};queue<x>q;  int N,K;int bfs(){   x a;    a.loc=N;    a.step=0;    q.push(a);    vis[N]=1;    while(!q.empty())    {        x t=q.front();        q.pop();        if(K==t.loc)        {            return t.step;        }        else        {            x temp;            if(t.loc-1>=0&&!vis[t.loc-1])            {                temp.loc=t.loc-1;                temp.step=t.step+1;                q.push(temp);                vis[temp.loc]=1;            }            if(t.loc+1<=100010&&!vis[t.loc+1])            {                temp.loc=t.loc+1;                temp.step=t.step+1;                q.push(temp);                vis[temp.loc+1]=1;            }            if(t.loc*2<=100010&&!vis[t.loc*2])            {                temp.loc=t.loc*2;                temp.step=t.step+1;                q.push(temp);                vis[temp.loc*2]=1;            }        }    }    return -1;}int main(){    int i,j,k;    int temp;    cin>>N>>K;    memset(vis,0,sizeof(vis));    cout<<bfs()<<endl;    return 0;}