POJ 3278 Catch That Cow

Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 30437 Accepted: 9390
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
这个看完了我顿时沸腾了起来，写的太好了，bfs与队列的完美结合，谢谢作者。
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#include <iostream>
#include <queue>
#define SIZE 100001

using namespace std;

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

int bfs(int n, int k)
{
    int head, next;
    //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
        head = x.front();
        //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
                step[next] = step[head] + 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;
}
作者：princeyuaner

补充：软件开发 , C++ ,