Catch That Cow

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

题意：有一个农民和一头牛，他们在一个数轴上，牛在k位置保持不动，农户开始时在n位置。设农户当前在M位置，每次移动时有三种选择：1.移动到M-1；2.移动到M+1位置；3.移动到M*2的位置。问最少移动多少次可以移动到牛所在的位置。所以可以用广搜来搜索这三个状态，直到搜索到牛所在的位置。

#include <iostream>#include <cstdio>#include <queue>#include <cstdlib>#include <cstring>using namespace std;int vis[200010],i;struct node{  int step,data;}s,v;int bfs(int x,int k){  queue<struct node>q;  s.data=x;  s.step=0;  q.push(s);  while(!q.empty())  {        v=q.front();        q.pop();        if(v.data==k)        {           printf("%d\n",v.step);           return 1;        }        if(v.data>=1&&!vis[v.data-1])        {           s.step=v.step+1;           s.data=v.data-1;           vis[v.data-1]=1;           q.push(s);        }        if(v.data<=k&&!vis[v.data+1])        {           s.step=v.step+1;           s.data=v.data+1;           vis[v.data+1]=1;           q.push(s);        }        if(v.data<=k&&!vis[v.data*2])        {           s.step=v.step+1;           s.data=v.data*2;           vis[v.data*2]=1;           q.push(s);        }  }  return -1;}int main(){    int n,k;    while(~scanf("%d%d",&n,&k))    {    memset(vis,0,sizeof(vis));    vis[n]=1;    bfs(n,k);    }    return 0;}

//此题与最短步数那个题一样最短步数用BFS中的队列分层存取各种状态，这个队列中第一层是走一步的step=1，位置有三个，每一层所有的位置全部进入队列该层只要有到达的点此时的step就是满足条件的步数，步数一步一步依次加一，