HDU 2717 Catch That Cow （bfs）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2717

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12615    Accepted Submission(s): 3902

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.

 

题目大意：在一条笔直的道路上，有一个农夫在A位置，一头牛在B位置，农夫一次可以搜索三个位置（例如农夫在x位置，他可以搜索x-1、x+1、x*2）问农夫至少需要几步能够找到牛。

解题思路： 用广搜 ，定义一个位置数组，每次搜索三个位置即可。

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
using namespace std;
int f[200020];  //记录步数 
void bfs(int n,int k)
{
    int a[3];  //位置数组 
    memset(f,0,sizeof(f));
    queue <int > q;
    q.push(n);
    f[q.front()] = 1;
    if (n == k)
        return ;
    while (!q.empty())
    {
        int t = q.front();
        int x = f[t];
        a[0] = t-1;  //三个位置 
        a[1] = t+1;
        a[2] = t*2;
        for (int i = 0; i < 3; i ++)
        {
            if (a[i] >= 0 && a[i] < 200001 && f[a[i]]==0)  //注意这里的边界值 
            {
                q.push(a[i]);
                f[a[i]] = x+1;  //在上一步的基础上加1 
            }
            if (a[i] == k)
                return ;
        }
        q.pop();
    }
}
int main ()
{
    int i;
    int n,k;
    while (~scanf("%d%d",&n,&k))
    {
        dfs(n,k);
        printf("%d
",f[k]-1); //减去农夫本来在的位置那一步 
    }
    return 0;
}