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

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[1000010];

int BFS(int x)
{
    int tx;
    now ans,tp;
    queue<now> que;
    ans.x=x,ans.step=0;
    vis[x]=true;
    que.push(ans);
    while(!que.empty())
    {
        tp=que.front();
        que.pop();
        tx=tp.x+1;   //no.1
        if(tx<0||tx>1000000);
        else
        {
            if(!vis[tx])
            {
                vis[tx]=true;
                ans.x=tx,ans.step=tp.step+1,que.push(ans);
                if(ans.x==K) return ans.step;
            }
        }
        tx=tp.x-1;    //no.2
        if(tx<0||tx>1000000);
        else
        {
            if(!vis[tx])
            {
                vis[tx]=true;
                ans.x=tx,ans.step=tp.step+1,que.push(ans);
                if(ans.x==K) return ans.step;
            }
        }
        tx=tp.x*2;   //no.3
        if(tx<0||tx>1000000);
        else
        {
            if(!vis[tx])
            {
                vis[tx]=true;
                ans.x=tx,ans.step=tp.step+1,que.push(ans);
                if(ans.x==K) return ans.step;
            }
        }
    }
}

int main()
{
    while(cin>>N>>K)
    {
        if(N==K)
        {
            cout<<"0"<<endl;
            continue;
        }
        memset(vis,false,sizeof(vis));
        cout<<BFS(N)<<endl;
    }
    return 0;
}

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。