POJ 3278 Catch That Cow（入门题目）

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

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.

Source
USACO 2007 Open Silver

题目大意：
FJ要抓奶牛。开始输入N（FJ的位置）K（奶牛的位置）。
FJ有三种移动方法：
向前走一步，耗时一分钟。
向后走一步，耗时一分钟；
向前移动到当前位置的两倍N*2，耗时一分钟。
问FJ抓到奶牛的最少时间。PS：奶牛是不会动的。

解法：
BFS+单调队列

#include<iostream>#include<cstring>#include<queue>using namespace std;#define Max 100005int n,k,step[Max],head,next;bool visit[Max];queue<int> q;int bfs(){    q.push(n);    step[n]=0;    visit[n]=1;    while(!q.empty())    {        head=q.front();        q.pop();        for(int i=0;i<3;i++)                      //三种选择：加1，减1，或者乘2         {            if(i==0)  next=head-1;            else if(i==1)  next=head+1;            else  next=head*2;             if(next>Max||next<0)                 //边界              continue;            if(!visit[next])            {                q.push(next);                step[next]=step[head]+1;                visit[next]=1;            }            if(next==k)                         //搜索跳出边界               return step[next];        }       }    return -1;}int main(){       while(cin>>n>>k)    {           memset(visit,0,sizeof(visit));        if(n>=k)          cout<<n-k<<endl;        else          cout<<bfs()<<endl;    }    return 0;}