hdu 2717 Catch That Cow(广搜bfs)

题目链接：http://i.cnblogs.com/EditPosts.aspx?opt=1

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7909    Accepted Submission(s): 2498

Problem 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.

 

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题目大意：在一个数轴上有n和k，农夫在n这个位置，奶牛在k那个位置，农夫要抓住奶牛，有两种方法：1、walking即农夫可以走x+1的位置和x-1的位置。

2、teleporting即每分钟可以走到2*x的位置。利用这两种方法，找到最快几步可以到达！

 

解题思路：其实就这三种情况，本来没打算用搜索的，不过发现好多东西都可以灵活运用！我们吧第一种方法看成两个方向，用dir[2]={1,-1,}表示，然后进行广搜！第二种方法自然就要特殊判断一下了！有一个地方要特殊判断一下，就是因为数据比较大，vis[ss.x]可能会超范围导致RE，所以都加一行判断使其保证在题目范围中。

 

详见代码。

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>

using namespace std;

int dir[3]= {1,-1};

struct node
{
    int x,step;
} s,ss;

int bfs(int n,int k)
{
    queue<node>q,qq;
    s.x=n;
    s.step=0;
    int vis[100010]= {0};
    q.push(s);
    while (!q.empty())
    {
        s=q.front();
        q.pop();
        if (s.x==k)
            return s.step;
        for (int i=0; i<2; i++)
        {
            ss.x=s.x+dir[i];
            ss.step=s.step+1;
            if (ss.x>=0&&ss.x<=100000)
                if (!vis[ss.x])
                {
                    vis[ss.x]=1;
                    q.push(ss);
                }
        }
        ss.x=s.x*2;
        ss.step=s.step+1;
        if (ss.x>=0&&ss.x<=100000)
        {
            if (!vis[ss.x])
            {
                vis[ss.x]=1;
                q.push(ss);
            }
        }
    }
    return 0;
}

int main ()
{
    int n,k;
    while (~scanf("%d%d",&n,&k))
    {
        printf ("%d
",bfs(n,k));
    }
    return 0;
}