USACO抓牛catchcow （bfs）

时间：2014-06-19 00:58:53 阅读：510 评论：0 收藏：0 [点我收藏+]

这题是黄巨大出的比赛题.

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

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

黄巨大的翻译很不错啊：

Problem 1 抓牛(catchcow.cpp/c/pas)

【题目描述】

农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上出发，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x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？

【输入格式】

仅有两个整数N和K

【输出格式】

最短时间

【样例输入】

5 17

【样例输出】

思路：1.正常人都想到的bfs，加上适当剪枝就能过

2.吴桐想的一种两次的dp（多次逼近最优解）

3.我刚看到题目想的2进制

还是先贴正常代码吧：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
using namespace std;
int d[200005];             //搜寻队列 
int f[200005];             //记录每个点最小值 
int head,tail,catchmax;
int main()
{
    int n,k,i,j,t;
    for (i=1;i<=100000;i++) f[i]=100000;
    cin>>n>>k;
    if (n>k) {cout<<n-k<<endl; return 0;}  //就怕丧心病狂倒着走的出题人  
    tail=1;
    head=1;
    d[tail]=n;
    f[n]=0;
    f[k]=k-n;                       //正着走最坏情况就是一步一步走过去，所以最大只会到k-n  
    while (head<=tail)
    {
         t=f[d[head]];
         catchmax=f[k];             //当前最优解 
         if (f[d[head]]<catchmax)   //比当前最优解大当然不用找了 
         {
                
         if ((d[head]*2<=2*k+1)&&(t+1<f[d[head]*2]))     //不能大于k的两倍+1，可以证明那样一定不会是最优解 
           {
            f[d[head]*2]=t+1;
            d[tail+1]=d[head]*2;
            tail++;
           }  
         if ((d[head]+1<=2*k+1)&&(t+1<f[d[head]+1]))
           {
            f[d[head]+1]=t+1;
            d[tail+1]=d[head]+1;
            tail++;
           }
         if ((d[head]+1<=2*k+1)&&(d[head]-1>=1)&&(t+1<f[d[head]-1]))
           {
            f[d[head]-1]=t+1;
            d[tail+1]=d[head]-1;
            tail++;    
           }
         }      
         head++;       //当时我把这句话放到前一个if判断中去死循环了。。因为如果到某个点等于当前最优解就进不了if判断也退不出while循环 
    }
    cout<<f[k]<<endl;
    return 0;
}

Run ID	User	Problem	Result	Memory	Time	Language	Code Length	Submit Time
12976028	seekdreamer	3278	Accepted	1268K	16MS	C++	1209B	2014-06-15 16:12:10

下面有正解：似乎跑起来比我的快一点

#include<iostream>
#include<cstdio>
#include<cstring>
#define N 100001
using namespace std;
int n,k,T,q[N],d[N];
bool inq[N];
void bfs()
{
    memset(d,127,sizeof(d));
    int t=0,w=1;
    q[0]=n;inq[n]=1;d[n]=0;
    while(t<w)
    {
        int now=q[t];t++;if(t==N)t=0;
        if(now<T&&d[now+1]>d[now]+1)
        {
            d[now+1]=d[now]+1;
            if((now+1)==n)return;
            if(!inq[now+1])
            {inq[now+1]=1;q[w++]=now+1;if(w==N)w=0;}
            }
        if(now>0&&d[now-1]>d[now]+1)
        {
            d[now-1]=d[now]+1;
            if((now-1)==n)return;
            if(!inq[now-1])
            {inq[now-1]=1;q[w++]=now-1;if(w==N)w=0;}
            }
        if((now<<1)<=T&&d[now<<1]>d[now]+1)
        {
            d[now<<1]=d[now]+1;
            if((now<<1)==n)return;
            if(!inq[now<<1])
            {inq[now<<1]=1;q[w++]=(now<<1);if(w==N)w=0;}
            }
    } 
}
int main()
{
    scanf("%d%d",&n,&k);
    T=max(n,k)+1; 
    bfs();
    printf("%d",d[k]);
    return 0;
}