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HDU2717:Catch That Cow

2013-04-21 21:53 190 查看
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Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4293    Accepted Submission(s): 1382


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

HintThe 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
 

Recommend

teddy
 

=====================================算法分析=====================================
BFS+剪枝（注意数组不要越界）。



=======================================剪枝=======================================
设John的坐标为X[John]，牛的坐标为X[cow]。
       

一、移动方法A（坐标加一）的剪枝：



    一旦X[John]>X[cow]，再往右走就没有意义了，因此只有在X[John]＜X[cow]时才需使用移动方法A。

二、移动方法B（坐标翻倍）的剪枝：



    假设：John使用移动方法B后X[John]＞X[cow]（即当前有2X[John]＞X[cow]）。

    如果：1、John选择使用移动方法B，则其最小移动次数显然为1＋2X[John]－X[cow]。

          2、John放弃使用移动方法B，则其最小移动次数不会超过X[cow]－X[John]。

    那么：需要选择移动方法B的前提是1＋2X[John]－X[cow]＜X[cow]－X[John]即3X[John]≤2X[cow]。

    又因：2X[John]≤X[cow]的情况下3X[John]≤2X[cow]是显然成立的。

    因此：只有在3X[John]≤2X[cow]时才需使用移动方法B。



=======================================代码=======================================


#include<queue>
#include<cstdio>
#include<cstring>

using namespace std;

const int MAXC=100000;

int N,K,Step[MAXC+5];

int BFS()
{
if(N==K) return 0;
memset(Step,-1,sizeof(Step));
Step
=0;
queue<int>q;
for(q.push(N);!q.empty();q.pop())
{
int cur=q.front();
for(int n=0;n<3;++n)
{
int tmp=cur;
if(n==0)
{
if(tmp*2<=MAXC&&tmp*3<=K*2) tmp*=2;
else continue;
}
else if(n==1)
{
if(tmp+1<=K) ++tmp;
else continue;
}
else if(n==2)
{
if(tmp-1>=0) --tmp;
else continue;
}
if(Step[tmp]==-1)
{
Step[tmp]=Step[cur]+1;
if(tmp==K) return Step[tmp];
q.push(tmp);
}
}
}
}

int main()
{
while(scanf("%d%d",&N,&K)==2)
{
printf("%d\n",BFS());
}
return 0;
}

                                            

                                        

                        
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