BZOJ1646: [Usaco2007 Open]Catch That Cow 抓住那隻牛

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?

    農夫約翰被通知，他的一隻奶牛逃逸了！所以他決定，馬上幽發，儘快把那隻奶牛抓回來．

    他們都站在數軸上．約翰在N(O≤N≤100000)處，奶牛在K(O≤K≤100000)處．約翰有

兩種辦法移動，步行和瞬移：步行每秒種可以讓約翰從z處走到x+l或x-l處；而瞬移則可讓他在1秒內從x處消失，在2x處出現．然而那隻逃逸的奶牛，悲劇地沒有發現自己的處境多麼糟糕，正站在那兒一動不動．

    那麼，約翰需要多少時間抓住那隻牛呢？

Input

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

    僅有兩個整數N和K.

Output

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

    最短的時間．

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.

題解：

  寬搜，判斷是否出界以及是否被使用過（因爲再次走到已經使用過的點一定不是最優解），然後把滿足條件的店推進隊列裏。

 

#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;
}