poj 3278 Catch That Cow If I know what love is, it is because of you

本文介绍了一个有趣的算法问题——如何让农夫最快抓到不动的逃逸奶牛。使用广度优先搜索(BFS)策略，通过步行和传送两种方式在数轴上移动。文章提供了完整的C++实现代码，并讨论了内存分配的影响。

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 39389		Accepted: 12263

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

这个题目我觉得如果叫我做是想不到的，用BFS，把所有的数想成是点，然后用各种办法去走，但是我RE了12次，后来我疯狂了，把原先的mark改成visit竟然过了，难以想象啊，大家说开到20w才可以过，看来也不是这样，我看人家开导10W+5,就过了嘛！！！！！

#include<iostream>
#include<queue>
#include<cstdio>
#include <memory.h>
#define MAX 100005
using namespace std;
queue<int> q;
bool visit[MAX];
int step[MAX];
int k,n;
int  bfs()
{

    //memset(step,0,sizeof(step));
    //memset(mark,0,sizeof(mark));
    int next,fis;
    q.push(n);
    visit[n]=1;
    step[n]=0;
    int i;
    while(!q.empty())
    {
        fis=q.front();
        q.pop();
        for(i=0; i<3; i++)
        {
            if(i==0)
                next=fis-1;
            else if(i==1)
                next=fis+1;
            else
               next=fis*2;
            if((next>MAX)||(next<0))
                continue;
            if(!visit[next])
            {
                q.push(next);
                step[next]=step[fis]+1;
                visit[next]=1;

            }
            if(next==k)
            {
                return step[next];

            }

        }

    }

}
int main()
{


     scanf("%ld%ld",&n,&k);
    memset(visit,0,sizeof(visit));
    if(n>=k)
        cout<<n-k<<endl;
    else
        cout<<bfs()<<endl;

    return 0;
}