poj-3278 catch that cow（搜索题）

题目描述：

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.

 

题目大意：人追牛，人可以加减一步或者两倍

解题思路：就是bfs搜索

 

 

AC代码

#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
using namespace std;
#define PI 3.14159265358979323846264338327950

struct point
{
    int x,y,step;
}st;

queue<point>q;
int vis[];
int n,m,flat;

int bfs()
{
    while(!q.empty())
    {
        q.pop();
    }

    memset(vis,,sizeof(vis));
    vis[st.x]=;
    q.push(st);
    while(!q.empty())
    {
        point now=q.front();
        if(now.x==m)
            return now.step;
        q.pop();
        for(int j=;j<;j++)
        {
            point next = now;
            if(j == )
                next.x=next.x+;
            else if(j==)
                next.x=next.x-;
            else
                next.x=next.x*;
            ++next.step;
            if(next.x==m)
                return next.step;
            if(next.x>= && next.x<= && !vis[next.x])
            {
                vis[next.x]=;
                q.push(next);
            }
        }
    }
    return ;
}
int main()
{
    while (~scanf("%d %d", &n, &m))
    {
        st.x = n;
        st.step=;
        printf("%d\n", bfs());
    }
  return ;

}