POJ 3278 Catch that cow

Catch That Cow
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 87429 Accepted: 27431
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

题意：在一个数轴上，有一个农民位于 n的位置处，有一头牛位于k 的位置处，农民 有三种走路方式：①若农民位于x ，农民可以移动一步到x+1 或x-1 ②若农民位 于x ，农民可以跳跃到2*x 处。问：农民需要最少多少步抓住那头牛？

题解：从农民的起点开始BFS广搜，通过农民的三种 移动方式(x-1，x+1，2*x)来向队列中插入节点，BFS广搜到牛的位置即可。

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;

const int maxn=100001;

bool vis[maxn];//标记数组
int step[maxn];//所走的步数
queue <int> q;

int bfs(int n,int k)
{
    int head,next;
    q.push(n);   //开始在n位置，n入队
    step[n]=0;
    vis[n]=true; //标记
    while(!q.empty())  //当队列非空
    {
        head=q.front();  //取队首
        q.pop();         //弹出对首
        for(int i=0;i<3;i++)     //三种走法
        {
            if(i==0) next=head-1;
            else if(i==1) next=head+1;
            else next=head*2;
            if(next<0 || next>=maxn) continue; //剪枝
            if(!vis[next])  //如果next位置未被访问
            {
                q.push(next);    //入队
                step[next]=step[head]+1;  //步数+1
                vis[next]=true;  //标记已访问
            }
            if(next==k) return step[next];  //遍历到结果，返回步数
        }
    }
}
int main()
{
    int n,k;
    while(cin>>n>>k)
    {
        memset(step,0,sizeof(step));
        memset(vis,false,sizeof(vis));

        while(!q.empty()) q.pop(); 
        if(n>=k) cout<<n-k<<endl;
        else cout<<bfs(n,k)<<endl;
    }
    return 0;
}