(简单) POJ 1278 Catch That Cow，回溯。

　　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?

　　也是典型的BFS的题目，我第一次设的数组最大是100000＊10，然后过了，第二次100000+5，居然也过了。。。

　　这个题就是从出发点开始bfs，有三条路可以走，一知道发现牛为止。

代码如下：

#include<iostream>
#include<cstring>
#include<queue>

using namespace std;

int N,K;
int rem[100005];

void bfs()
{
    queue <int> que;
    int t;

    que.push(N);
    rem[N]=0;
    
    while(!que.empty())
    {
        t=que.front();
        que.pop();

        if(t==K)
            return;

        if(t*2<=100000&&rem[t*2]==-1)
        {
            rem[t*2]=rem[t]+1;
            que.push(t*2);
        }
        if(t+1<=100000&&rem[t+1]==-1)
        {
            rem[t+1]=rem[t]+1;
            que.push(t+1);
        }
        if(t-1>=0&&rem[t-1]==-1)
        {
            rem[t-1]=rem[t]+1;
            que.push(t-1);
        }
    }
}

int main()
{
    ios::sync_with_stdio(false);

    while(cin>>N>>K)
    {
        memset(rem,-1,sizeof(rem));

        if(K<=N)
            cout<<N-K<<endl;
        else
        {
            bfs();
            cout<<rem[K]<<endl;
        }
    }

    return 0;
}