BFS --- 模板题

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36079		Accepted: 11123

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.

【题目来源】

USACO 2007 Open Silver

【题目大意】

在x轴的正半轴上，一个人的出发点是N,一头牛在另一点K，人可以有两种操作：

1.步行：+1 or -1

2.传送：*2

问你通过最少的步数到达牛的位置，需要多少步。

【题目分析】

就是一个裸的广搜，使用队列实现。

#include<iostream>

#include<cstring>

#include<queue>

using namespace std;

int N,K;

struct now

{

    int x;

    int step;

};

bool vis[];

int BFS(int x)

{

    int tx;

    now ans,tp;

    queue<now> que;

    ans.x=x,ans.step=;

    vis[x]=true;

    que.push(ans);

    while(!que.empty())

    {

        tp=que.front();

        que.pop();

        tx=tp.x+;   //no.1

        if(tx<||tx>);

        else

        {

            if(!vis[tx])

            {

                vis[tx]=true;

                ans.x=tx,ans.step=tp.step+,que.push(ans);

                if(ans.x==K) return ans.step;

            }

        }

        tx=tp.x-;    //no.2

        if(tx<||tx>);

        else

        {

            if(!vis[tx])

            {

                vis[tx]=true;

                ans.x=tx,ans.step=tp.step+,que.push(ans);

                if(ans.x==K) return ans.step;

            }

        }

        tx=tp.x*;   //no.3

        if(tx<||tx>);

        else

        {

            if(!vis[tx])

            {

                vis[tx]=true;

                ans.x=tx,ans.step=tp.step+,que.push(ans);

                if(ans.x==K) return ans.step;

            }

        }

    }

}

int main()

{

    while(cin>>N>>K)

    {

        if(N==K)

        {

            cout<<""<<endl;

            continue;

        }

        memset(vis,false,sizeof(vis));

        cout<<BFS(N)<<endl;

    }

    return ;

}

当然还可以剪枝一下，比如说：如果人的坐标大于牛的坐标，这时就只需要做-1这一步就可以了。

巴特西

BFS --- 模板题

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