catch the cow(简单题)

题目链接：点击打开链接

代码：

#include
#include
#include
using namespace std;
#include
struct  qq
{
    int x,y;
} w;
queueq;
int k;
int e[200010];
void  dfs(int n)
{
    w.x=n;
    w.y=0;
    while(!q.empty())
    {
        q.pop();
    }
    q.push(w);
    while(!q.empty())
    {
        qq n1=q.front();
        q.pop();
        for(int i=0; i<3; i++)
        {
            qq n2;
            if(i==0)
                n2.x=n1.x*2;
            else  if(i==1)
                n2.x=n1.x-1;
            else
                n2.x=n1.x+1;
            if(n2.x<0||n2.x>100000||e[n2.x])
                continue;
            n2.y=n1.y+1;
            e[n2.x]=1;
            if(n2.x==k)
            {
                printf("%d\n",n2.y);
                return ;
            }
            q.push(n2);
        }
    }
    return ;
}
int main()
{
    int n;
    scanf("%d%d",&n,&k);
    if(n==k)
        printf("0\n");
    else if(n>k)
        printf("%d\n",n-k);
    else
    {
        memset(e,0,sizeof(e));
        e[n]=1;
        dfs(n);
    }
    return 0;
}

简单题