# POJ 3278 Catch That Cow C++版

POJ题解 4995浏览

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: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Output

For each k given in the input, print one line containing ak to the output.

Sample Input

5 17

Sample Output

4

Slyar:USACO的题，给出2个数n和k，问从n经过+1或者-1或者*2能到达k的最小步数。分3个方向BFS，找到后树的深度就是最小步数了。注意n可以比k大，这时只有-1一种办法可以从n到达k，直接减就行了，还有要注意边界的判断…

```#include
#include
#define SIZE 100001

using namespace std;

queue x;
bool visited[SIZE];
int step[SIZE];

int bfs(int n, int k)
{
//起始节点入队
x.push(n);
//标记n已访问
visited[n] = true;
//起始步数为0
step[n] = 0;
//队列非空时
while (!x.empty())
{
//取出队头
//弹出队头
x.pop();
//3个方向搜索
for (int i = 0; i < 3; i++)
{
if (i == 0) next = head - 1;
else if (i == 1) next = head + 1;
else next = head * 2;
//越界就不考虑了
if (next > SIZE || next < 0) continue;
//判重
if (!visited[next])
{
//节点入队
x.push(next);
//步数+1
//标记节点已访问
visited[next] = true;
}
//找到退出
if (next == k) return step[next];
}
}
}

int main()
{
int n, k;
cin >> n >> k;
if (n >= k)
{
cout << n - k << endl;
}
else
{
cout << bfs(n, k) << endl;
}
return 0;
}
```

### 网友最新评论 (3)

1. 哈哈哈，从你这学到好多东西啊，非常感谢
蜗牛10年前 (2011-04-10)回复
2. 好久没看C++了，当初只学了一点皮毛。
网络人生11年前 (2010-04-18)回复