POJ3278 Catch That Cow(BFS)

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 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

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.

在本题中，农场主有三种方式，一种是往前走一步，一种是往后走一步，还有一种是瞬移到原来位置的两倍。

在BFS循环的时候就可以将这三种

操作进行处理。走过的位置标记。代表不会走回原来的位置。因为不标记的话，会导致死循环.

在找到牛的时候跳出。

代码：

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;

const int maxn = 100000+5;

int n,k;
int vis[maxn];

struct point
{
    int x,step; //坐标和时间
};

int check(int x)//判断是否满足条件
{
    if(x>=0&&x<=100000 && vis[x]==0)
        return 1;
    return 0;
}

int bfs()
{
    memset(vis, 0, sizeof(vis));
    queue< point > q;
    point st,now;
    //将起点加入队列
    st.x = n;
    st.step = 0;
    vis[st.x] = 1; // 记录已经访问
    q.push(st);

    while(!q.empty())
    {
        now = q.front();
        q.pop();

        if(now.x == k) //结束条件
        {
            return now.step;
        }

        //循环三种情况
        for(int j = 0 ; j < 3; j++)
        {
            point next = now;//为了复制now的step

            if(j == 0)
            {
                next.x = now.x + 1;
            }
            else if(j == 1)
            {
                next.x = now.x - 1;
            }
            else if(j == 2)
            {
                next.x = 2 * now.x;
            }

            if(check(next.x)) //如果这个点满足条件
            {
                next.step++;
                vis[next.x] = 1;
                q.push(next);
            }
        }
    }
    return -1;
}


int main()
{
    scanf("%d%d",&n,&k);
    printf("%d\n",bfs());
}

POJ3278 Catch That Cow(BFS)

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