POJ 3278 Catch That Cow

[align=center]Catch That Cow[/align]

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 37605		Accepted: 11656

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
 
题意：
给出农夫和牛的位置 农夫每次只能向前或向后走一步或向前走2*x(x为农夫当前位置)步
求农夫抓到牛的最短时间
 
代码：

#include <iostream>
#include <queue>
# include <cstring>
#include<cstdio>
#define SIZE 100001

using namespace std;
queue<int> x;
int step[SIZE];
char visited[SIZE]={0};
int bfs(int n, int k)
{
memset (visited,0,SIZE);
while (!x.empty())		//队列为非空时
x.pop();		//从队列前面移除元素，即将队列置为空
int head, next;
x.push(n);		//起始节点入队
visited
= true;		//标记n已访问
step
= 0;		//起始步数为0
while (!x.empty())		//队列为非空时
{
head = x.front();		//取出队头
x.pop();		//移除队头
for (int i = 0; i < 3; i++)		//三个方向搜索
{
if (i == 0) next = head - 1;
else if (i == 1) next = head + 1;
else next = head * 2;		//三种情况

if (next > SIZE || next < 0) continue;		//越界就不考虑了

if (!visited[next])		//判重，若访问过则不用再访问
{
x.push(next);		//节点入队
step[next] = step[head] + 1;		//步数+1
visited[next] = 1;		//标记节点已访问
}

if (next == k) return step[next];		//找到退出
}
}
}

int main()
{
int n, k;
while (scanf ("%d%d",&n,&k)!=EOF)
{
if (n >= k)
cout << n - k << endl;		//若n>=k,只能通过x-1操作
else
cout << bfs(n, k) << endl;
}
return 0;
}

思路：
bfs即可