[POJ](3278)Catch That Cow ---BFS（图）

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 96514		Accepted: 30289

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.

题意：John要去抓他的奶牛，现在John在n位置，奶牛在k位置。奶牛位置不变，

John有三种方式去其他位置 :

①：向后走一步，耗时一分钟。

②：向前走一步，耗时一分钟。

③：到达John所处位置的两倍位置，耗时一分钟。

问：John如何用时最少找到他的奶牛？

思路：将问题抽象为图，John在某一位置N，有三种位置可走，N-1，N+1，2*N，即任一位置都有与该位置连通的三个邻接点。采用BFS的算法，不断搜索位置，先到位置k的路径即为耗时最少的。

AC代码: 

注:代码在POJ上用C++交AC，G++交WA。可能是编译器优化的原因( ⊙ o ⊙ )

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int maxn=100005;
bool vis[maxn];
int sum[maxn];
int bfs(int n,int k)
{
    queue<int>q;
    vis[n]=true;
    sum[n]=0;
    q.push(n);
    int next;
    while(!q.empty())
    {
        int now=q.front();
        q.pop();
        for(int i=1;i<=3;i++) //每个点有三种走法
        {
            if(i==1)
                next=now-1; //向后走一步
            else if(i==2)
                next=now+1; //向前走一步
            else if(i==3)
                next=2*now; //向前走到当前位置的两倍位置
            if(next<0 || next>maxn)  //防止越界
                continue;
            if(!vis[next])
            {
                vis[next]=true;
                sum[next]=sum[now]+1;
                q.push(next);
            }
            if(next==k) //先到达奶牛位置的肯定是用时最短的
                return sum[next];
        }
    }

}
int main()
{
    int n,k;
    while(cin>>n>>k)
    {
        memset(vis,false,sizeof(vis));
        memset(sum,0,sizeof(sum));
        if(n>=k)
            cout<<n-k<<endl;
        else
            cout<<bfs(n,k)<<endl;
    }
    return 0;
}