POJ3278 农夫追牛（BFS

最新推荐文章于 2020-04-15 10:29:40 发布

原创最新推荐文章于 2020-04-15 10:29:40 发布 · 1k 阅读

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本文介绍了一道经典的算法题目，即如何利用最少步骤从给定起点捕捉到静止目标。通过对三种移动方式的分析，采用广度优先搜索(BFS)策略解决了问题，并对比了深度优先搜索(DFS)的应用场景。

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上题目先：

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

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.

题目大概意思是给你农夫其实位置（一维），然后另一个数是奶牛的坐标，就是问你农夫有三种方式

1：向前走1（+1

2：向后走1（-1

3：坐车做到当前位置的2倍（*2

上面这三种耗时都是1，问最快什么时候抓住奶牛（奶牛不动的

最初我以为这题是DFS的真的和白书DFS模板题太像了也一直在想怎么用DFS来过，但是后来真的只能做到判断，实在无法求出他的最佳耗时，然后就在网上看题解都是用BFS做的，然后就寻思这把BFS学了在做，现在来看这个问题也明朗了很多，还是通过队列来保存最优解的分步。

主要是给了我一个关于DFS与BFS的教训：求最优解（最短路径。最少操作应该往BFS想，而想白书模板那种问是否存在的才真正应该用DFS解决

AC截图：

AC CODE:

/*
Sample Input
5 17
Sample Output
4
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn=100010;
int vis[maxn+6];
int n,m;
struct cnmb
{
  int x,step;
};
queue<cnmb>q;
int main()
{
  scanf("%d%d",&n,&m);
  memset(vis,0,sizeof(vis));
  cnmb wmw;
  wmw.x=n;
  wmw.step=0;
  q.push(wmw);
  vis[n]=1;
  while(!q.empty()){
    cnmb wyh=q.front();
    if(wyh.x==m){
      printf("%d\n",wyh.step);
      return 0;
    }
    if(wyh.x-1>=0&&!vis[wyh.x-1])
    {
      cnmb lxc;
      lxc.x=wyh.x-1;
      lxc.step=wyh.step+1;
      q.push(lxc);
      vis[wyh.x-1]=1;
    }
    if(wyh.x+1<maxn&&!vis[wyh.x+1])
    {
      cnmb xjw;
      xjw.x=wyh.x+1;
      xjw.step=wyh.step+1;
      q.push(xjw);
      vis[wyh.x+1]=1;
    }
    if(wyh.x*2<maxn&&!vis[wyh.x*2])
    {
      cnmb sb;
      sb.x=wyh.x*2;
      sb.step=wyh.step+1;
      q.push(sb);
      vis[wyh.x*2]=1;
    }
    q.pop();
  }
  //return 0;
}

刷题吧弱鸡！