poj 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 30924 Accepted: 9536

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

题目大意： 在一个坐标轴上给你点n和点k，设当前点位x，可以有三种方式，x+1或者x-1或者x*2，问最少几步可以打k点；

   直接bfs，搞个队列；

#include <iostream>#include<cstdio>#include<cstring>#include<queue>#define MAX 100005using namespace std;queue <int>q;bool visit[MAX];//标记该点是否走过的数组int step[MAX];//表示到当前点所走的步数的数组bool pd(int num)//判断当前点是否越界，搜索题必要步骤{    if(num<0||num>100000)        return true;      return false;}int bfs(int st,int ed){    queue<int>q;    int t,tmp;    q.push(st);//先将起点入队列    visit[st]=true;//将起点设置为已经走的    while(!q.empty())    {        t=q.front();        q.pop();        for(int i=0;i<3;i++)//对三种方式进行搜索        {            if(i==0)                tmp=t+1;            else if(i==1)                tmp=t-1;            else                tmp=t*2;            if(pd(tmp))//判断是否越界                continue;            if(!visit[tmp])            {                step[tmp]=step[t]+1;//当前点的步数为上一点+1                if(tmp==ed)//若已到达终点则 返回所走步数                    return step[tmp];                visit[tmp]=true;//标记tmp点已走过                q.push(tmp);            }        }    }}int main(){   int n,k;   while(~scanf("%d%d",&n,&k))   {       memset(visit,false,sizeof(visit));       if(n>=k)//如果起点大于终点，则只有一种方式可以选择即x-1        cout<<n-k<<endl;//所以直接输出两者差值       else        cout<<bfs(n,k)<<endl;   }    return 0;}