Catch That Cow POJ - 3278

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本文介绍了一种利用广度优先搜索(BFS)算法解决迷宫逃脱问题的方法，具体场景为农民John(FJ)如何在最短时间内追上静止不动的逃逸奶牛。通过在数轴上进行行走或瞬移操作，文章详细解释了算法实现过程，包括边界条件判断、状态转移和最优路径输出。

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思路：每到达一个位置有3种选择，分别是步行（+1）、步行（-1）、传送（*2），所以要用广度优先搜索bfs，把所有符合条件即的下一步路线（即不会在下一步就产生越界的路线）都加入到队列中作为候选。首先考虑一种特殊情况，即n>=k，此时只能后退，所以花费的时间为n-k；若n< k，则需要使用bfs来找出一条耗时最少的方案。

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.

#include <iostream>
#include<queue>
using namespace std;
int step[100002];
queue<int>q;
bool check(int length)
{
    if(length>=0&&length<=100000&&!step[length])
        return true;
        else
            return false;
}
void bfs(int n,int k)
{
    int s1,s2;
    q.push(n);    //当前坐标入列
    step[n]=1;    //当前坐标是路径中的第一步
    while(!q.empty())    //若队列不为空
    {
        s1=q.front();    //s1保存队首坐标
        if(s1==k)    //若刚好当前坐标等于k则输出当前步数（即时间），退出bfs
        {
            cout<<step[s1]-1<<endl;
            return;
        }
        q.pop();    //队首出列
        s2=s1;    //s2保存队首坐标
        if(check(s2+1))    //若n=n+1未被访问过，且没有越界
        {
            step[s2+1]=step[s2]+1;    //n=n+1作为路径的下一步候选
            q.push(s2+1);    //入列
        }
        s2=s1;
        if(check(s2-1))    //若n=n-1未被访问过，且没有越界
        {
            step[s2-1]=step[s2]+1;    //n=n-1作为路径的下一步候选
            q.push(s2-1);    //入列
        }
        s2=s1;
        if(check(s2*2))    //若n=n*2未被访问过，且没有越界
        {
            step[s2*2]=step[s2]+1;    //n=n*2作为路径的下一步候选
            q.push(s2*2);    //入列
        }

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