BZOJ1646: [Usaco2007 Open]Catch That Cow 抓住那只牛

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?

    农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来．

    他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有

两种办法移动，步行和瞬移：步行每秒种可以让约翰从z处走到x+l或x-l处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．

    那么，约翰需要多少时间抓住那只牛呢？

Input

* Line 1: Two space-separated integers: N and K

    仅有两个整数N和K.

Output

* Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

    最短的时间．

Sample Input

5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.

Sample Output

4

OUTPUT DETAILS:

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.

题解：

  宽搜，判断是否出界以及是否被使用过（因为再次走到已经使用过的点一定不是最优解），然后把满足条件的店推进队列里。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int MAXN=500001;
bool visit[MAXN];
vector <int> Q;
struct xx
{
	int s,cnt;
}a[MAXN];
int main(int argc, char *argv[])
{
	int n,k,head,tail,d=0,zt,i;
	scanf("%d%d",&n,&k);
	a[1].s=n;
	head=1,tail=1;
	while(head<=tail)
	{
		zt=tail;
		for(i=head;i<=tail;i++)
		{
			if(a[i].s==k) {
				printf("%d\n",a[i].cnt);
				return 0;
			}
			if(a[i].s+1<MAXN)
			if(visit[a[i].s+1]==false) 
			{
				zt++;
				a[zt].s=a[i].s+1;
				a[zt].cnt=a[i].cnt+1;
				visit[a[zt].s]=true;
			}
			if(a[i].s-1>=0)
			if(visit[a[i].s-1]==false) 
			{
				zt++;
				a[zt].s=a[i].s-1;
				a[zt].cnt=a[i].cnt+1;
				visit[a[zt].s]=true;
			}
			if(a[i].s*2<MAXN)
			if(visit[a[i].s*2]==false) 
			{
				zt++;
				a[zt].s=a[i].s*2;
				a[zt].cnt=a[i].cnt+1;
				visit[a[zt].s]=true;
			}
			head=tail+1;
			tail=zt;
		}
	}
	return 0;
}