Catch That Cow（BFS广搜）

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.

 

 

广搜的基本例题：

 

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
using namespace std;
struct point
{
    int x;///记录位置
    int count;///记录步数
};
queue<point>q;
struct point s,now,t;
int vis[200000];///假设FJ开始的位置就是100000，那么变化两倍之后就是200000
int bfs(int n,int m)
{
    int j;
    while(!q.empty())
    {
        q.pop();
    }///清空队列
    memset(vis,0,sizeof(vis));
    vis[s.x]=1;
    q.push(s);
    while(!q.empty())
    {
        t=q.front();
        if(t.x==m)
            return t.count;
        for(j=0; j<3; j++)
        {
            now=t;
            if(j==0)
            {
                now.x=now.x+1;
            }
            else if (j==1)
            {
                now.x=now.x-1;
            }
            else if(j==2)
            {
                now.x=now.x*2;
            }
            now.count++;
            if(now.x==m)
            {
                return now.count;
            }
            if(now.x>=0&&now.x<=200000&&vis[now.x]==0)
            {
                vis[now.x]=1;
                q.push(now);
            }
        }
        q.pop();
    }
    return 0;///二者开始的位置相同
}
int main()
{
    int n,m,ans;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        s.x=n;
        s.count=0;
        ans=bfs(n,m);
        printf("%d
",ans);
    }
    return 0;
}

反思：这道题和之前的那一道剑客救公主那一道题一样，不仅仅需要考虑题意之中的搜索方式，还要考虑搜索不到或者起始位置与终止位置相同等特殊情况，该去如何设置被调函数，该去返回一个什么样的值，这两道题都是因为这一点使我wa了好多次，引以为戒。