Catch That Cow POJ - 3278

Catch That Cow POJ - 3278

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.

大致题意
农夫John要抓回一头牛，他们位置都在x轴上，在单位时间John可以前进一步x+1或者后退一步x-1，或者前进x步即2*x，现在分别告诉你John的位置和牛的位置，问需要抓到牛的最少时间。

思路
bfs求最优解

下面是代码

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <fstream>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <queue>#include <stack>#include <vector>#include <list>#include<sstream>#include<ctime>using namespace std;  struct node{    int x,step;   //定义结构体node记录坐标x和已用单位时间};int n,k;queue<node> Q;   //bfs所用队列int f[200005];   //用一维数组f来记录某坐标点是否已经走过void bfs()  {        int x,step;      while(!Q.empty())      {          node tmp=Q.front();   //取队首元素          Q.pop();              //出队列          x=tmp.x;          step=tmp.step;          if(x==k)          //已经到达牛所在的位置，结束bfs，输出结果          {              cout<<step;              return;            }         if(x>=1&&!f[x-1])     //边界判断，判断该位置是否已经走过          {             node temp;                    f[x-1]=1;             temp.x=x-1;             temp.step=step+1;             Q.push(temp);           }           if(x<=k&&!f[x+1])            {             node temp;             f[x+1]=1;             temp.x=x+1;             temp.step=step+1;             Q.push(temp);             }             if(x<=k&&!f[x*2])             {                node temp;                f[x*2]=1;                temp.x=2*x;                temp.step=step+1;                Q.push(temp);                 }        }     }  int main()  {     cin>>n>>k;   memset(f,0,sizeof(int)); //初始化   f[n]=1;   node a;   a.x=n;   a.step=0;   Q.push(a);   bfs();    return 0;  }