POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 35330		Accepted: 10888

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.

 

昨天写的BFS一直不对，今天早上仔细检查了一下，发现是自己剪枝没处理好，把有最少步骤的枝剪掉了，囧……

之前写了一个在读入n以后，如果2*n<k就自动n*=2，后来发现这样并不是最快的，很有可能最后剩下一个很长的路只能一步一步+1

所以“退一步海阔天空”很重要~~

 

下面是我写的代码（第一次做BFS，之前只在刘汝佳的书上学过一次BFS）

 

#include<iostream>
#include<queue>
#include<cstdio>

using namespace std;

long step[100001];
long n,k;
queue<int> q;

int main()
{
    cin>>n>>k;
    if(n>k)
        cout<<n-k<<endl;
    else
    {
        q.push(n);
        while((!q.empty())&&q.front()!=k)
        {
            int z;
            //首先判断z的范围是不是越界，如果不越界再继续判断后面的条件：如果step[z]还没有被赋值过值（即step[z]==0），就把这次的步数赋进去（这里需要注意step[n]原本就是0），如果已经付过值且比这次的大，就把这个新算出来的小的赋进去
            if(((z=q.front()-1)>=1)&&((z!=n&&step[z]==0)||(step[z]>(step[q.front()]+1))))
            {
                step[z]=step[q.front()]+1;
                if(z==k)
                    break;
                else
                    q.push(z);
            }
            if(((z=q.front()+1)<=100000)&&((z!=n&&step[z]==0)||(step[z]>(step[q.front()]+1))))
            {
                step[z]=step[q.front()]+1;
                if(z==k)
                    break;
                else if(z<k)
                    q.push(z);
                else
                    for(int x=z-1,y=1;x>=k;x--,y++)
                        if((x!=n&&step[x]==0)||step[x]>step[z]+y)
                            step[x]=step[z]+y;
            }
            if(((z=q.front()*2)<=100000)&&((z!=n&&step[z]==0)||(step[z]>(step[q.front()]+1))))
            {
                step[z]=step[q.front()]+1;
                if(z==k)
                    break;
                else if(z<k)
                    q.push(z);
                else 
                {
                    //如果z>k就只能一步一步退回去了，所以不需要入队，只需一个一个退回去就能算出步骤
                    for(int x=z-1,y=1;x>=k;x--,y++)
                        if((x!=n&&step[x]==0)||step[x]>(step[z]+y))
                            step[x]=step[z]+y;
                }
            }
            q.pop();
        }

        cout<<step[k]<<endl;
    }

    return 0;
}