Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory limit: 65536K

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?

输入

Line 1: Two space-separated integers: N and K

输出

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

示例输入

5 17

示例输出

4

提示

//本题的意思大概是：FJ在N，奶牛在K，可以通过三种方法抓到奶牛。1.到现在位置的2倍的位置2.到现在位置的下一个位置3.到现在位置的前一个位置。使用方法BFS来做。

#include<bits/stdc++.h>
#include<queue>
using namespace std;
int n;

int k;
const int N=1000000;
int vis[N+10];
struct node
{
    int x,step;
};
int check(int x)
{
    if(x<0||x>=N||vis[x])
        return 0;
    else return 1;
}
int BFS(int x)
{
    queue<node>q;
    struct node a,next;
    a.x=x;
    a.step=0;
    vis[x]=1;
    q.push(a);
    while(!q.empty())
    {
        a=q.front();
        q.pop();
        if(a.x==k)
            return a.step;
           next.x=a.x+1;
        if(check(next.x))
        {
            next.step=a.step+1;     //操作的次数
            vis[next.x]=1;
            q.push(next);
        }
        next.x=a.x-1;
        if(check(next.x))
        {
            next.step=a.step+1;
            vis[next.x]=1;
            q.push(next);
        }
        next.x=a.x*2;
        if(check(next.x))
        {
            next.step=a.step+1;
            vis[next.x]=1;
            q.push(next);
        }
    }
    return -1;
}
int main()
{
        int o;
        while(~scanf("%d%d",&n,&k))
        {
        memset(vis,0,sizeof(vis));
        o=BFS(n);
        printf("%d\n",o);
        }
         return 0;
}