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;}