图的搜索方式-Catch that Cow (BFS）

 

 Hint: 此题可以看成是一颗三叉树进行层序遍历。如果把牛的坐标看成是不变的量，让农夫动，那么就会有三种情况：向左，向            右，还有除二操作。 注意用队列的时候要理解清楚谁进谁出

       

 Accepted;    代码1.0版（整体的思路不是很清晰）

#include 
#include 
#include 
int vj[100100];
int main()
{
    memset(vj,-1,sizeof(vj)); //用途有两个，第一是标记是否搜索过，第二记录走过的步数
    int n, k, head, end;
    int s[10010]; //队列数组
    scanf("%d %d",&n,&k);
    if(n > k)  //分两种情况，农夫在牛的前面，和在后面
        printf("%d\n",n-k);
    else
    {
        head = 0;
        end = 1;
        s[0] = k;
        vj[k] = 0;//表示已经走过了原始的k点，做标记
        while(head < end)
        {
            if(s[head] == 0)
            {
                if(vj[s[head]-1] == -1)
                {
                    s[end++] = s[head]-1;
                    vj[s[head]+1] = vj[s[head]]+1;//因为vj定义的是-1；所以要加一才能被标记
                }
            }
            if(vj[s[head]+1] == -1)
            {
                s[end++] = s[head]+1;
                vj[s[head]+1] = vj[s[head]]+1;
            }
            if(s[head]%2 == 0&&vj[s[head]/2] == -1)
            {
                s[end++] = s[head]/2;
                vj[s[head]/2] = vj[s[head]]+1;
            }
            if(vj[n] != -1)
                break;
            head++;
        }
        printf("%d\n",vj[n]);
    }
    return 0;
}

 

 Accepted;    代码2.0版（思路清晰）

 

#include 
#include 
#include 
int vj[100100], s[100100];
int head, end;
int main()
{
    int n, k, g;
    memset(vj,-1,sizeof(vj));
    head = end = 0;
    scanf("%d %d",&n,&k);
    end++;
    s[end] = k;
    if(n >= k)
        printf("%d\n",n-k);
    else
    {
        vj[k] = 0;
        while(head < end)
        {
            head++;
            g = s[head];
            if(vj[g+1] == -1)
            {
                vj[g+1] = vj[g]+1;
                end++;
                s[end] = g+1;
            }
            if(vj[g-1] == -1)
            {
                vj[g-1] = vj[g]+1;
                end++;
                s[end] = g-1;
            }
            if(g % 2==0&&vj[g/2] == -1)
            {
                vj[g/2] = vj[g]+1;
                end++;
                s[end] = g/2;
            }
        }
        printf("%d\n",vj[n]);
    }
    return 0;
}