【POJ】3278 - Catch That Cow（bfs，队列）

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 68324		Accepted: 21515

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

搜索变化数的时候，注意不要让数字越界（无论是越最大值还是越到-1）

代码如下：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAX 200222
using namespace std;
int n,k;
int map[MAX];
struct node
{
	int num,stp;
}a,b;
int bfs()
{
	queue<node>q;
	a.num = n;
	a.stp = 0;
	map[n] = 1;
	q.push(a);
	while (1)
	{
		a = q.front();
		q.pop();
		for (int i = 0;i < 3;i++)
		{
			if (i == 0)
			{
				b.num = a.num+1;
				if (b.num == k)
					return a.stp+1;
				if (b.num < MAX-200 && map[b.num] == 0)
				{
					b.stp = a.stp+1;
					map[b.num] = 1;
					q.push(b);
				}
			
			}
			else if (i == 1)
			{
				b.num = a.num-1;
				if (b.num == k)
					return b.stp;
				if (b.num < MAX-200 && b.num>=0 && map[b.num] == 0)		//这里要注意不要让num为-1 
				{
					b.stp = a.stp+1;
					map[b.num] = 1;
					q.push(b);
				}
			}
			else
			{
				b.num = a.num*2;
				if (b.num == k)
					return b.stp;
				if (b.num < MAX-200 && map[b.num] == 0)
				{
					b.stp = a.stp+1;
					map[b.num] = 1;
					q.push(b);
				}
			}
		}
	}
}
int main()
{
	int ans;
	while (~scanf ("%d %d",&n,&k))
	{
		memset (map,0,sizeof(map));
		if (n >= k)
			printf ("%d\n",n-k);
		else
		{
			ans = bfs();
			printf ("%d\n",ans);
		}
	}
	return 0;
}