Catch That Cow——广搜

 

Catch That Cow

 

 

Problem 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.[/hint]

 

 

 

题目大意：农夫的牛丢了，现在给农夫的位置N，和牛的位置K。一维坐标。现在假如牛不动，农夫去找牛，有两种走路方式，一种x-1，和x+1用时1。

 

一种2*x，用时1 。求最短时间。

 

简单广搜，搜索入门题。

这道题要标记找过的不能找了，超出范围的不能找了，否则会时间超限。

 

 

 

#include///Catch That Cow
using namespace std;
int to[2]={1,-1};
int a,b,sum;
int vis[100000];
struct place
{
    int x,time;
};
int check(place k)
{
    if(k.x<0||k.x>100000||vis[k.x]==1)
        return 0;
    return 1;
}
int bfs(place n)
{
    place m,next;
    queuew;
    w.push(n);
    while(!w.empty())
    {
        m=w.front();
        w.pop();
        if(m.x==b)
            return m.time;
        for(int i=0;i<2;i++)
        {
            next.x=m.x+to[i];
            next.time=m.time+1;
            if(next.x==b)
                return next.time;
            if(check(next))
            {
                w.push(next);
                vis[next.x]=1;
            }
        }
        next.x=m.x*2;
        next.time=m.time+1;
        if(next.x==b)
            return next.time;
        if(check(next))
        {
            w.push(next);
            vis[next.x]=1;
        }
    }
    return 0;
}
int main()
{
    int i,j,t;
    place x1;
    while(~scanf("%d %d",&a,&b))
    {
        memset(vis,0,sizeof(vis));
        x1.x=a;
        x1.time=0;
        vis[x1.x]=1;
        sum=0;
        sum=bfs(x1);
        printf("%d\n",sum);
    }
    return 0;
}