POJ 3278 Catch That Cow(DFS)

最新推荐文章于 2021-02-27 17:31:42 发布

wuhuanbin

最新推荐文章于 2021-02-27 17:31:42 发布

阅读量937

点赞数 3

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/SCAU_BIN/article/details/51482291

本篇介绍了一个有趣的算法问题，即如何利用行走和传送两种方式在最短时间内从起点N到达固定位置的终点K。通过递归算法实现最优路径的计算。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

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.

题解：分情况，设f(n,t) 是 n 到 t 的最短时间。
1.如果n>k，直接输出n-k；

2.如果n=0，那么走n-1和走2*n都是无意义的，只能走n+1，此时n先走到1，故f(0,k)=(1,k)+1，且令n=1往下操作；

3.如果k是偶数，a情况：如果k/2>n,f(n，k) = f(n，k/2)+1；
b情况：如果k/2<n,f(n，k) = min( f(n，k/2)+1，k-n(直接一步步过去) );

4.如果k是奇数，那么必定是f(n,k) = min(f(n，k+1)，f(n，k-1))+1;

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int  n,m;
int dfs(int cur,int target)
{

    if(cur>=target)
        return cur-target;
    if(target%2==0)
    {
        if(cur<target/2)
            return dfs(cur,target/2)+1;
        else if(cur==target/2)
            return 1;
        else
            return min(1+cur-target/2,target-cur);
    }
    else
    {
        return min(dfs(cur,target+1),dfs(cur,target-1))+1;
    }
}
int main()
{
    int flag=0;
    scanf("%d %d",&n,&m);
    if(n==0)
        flag=1;
    printf("%d\n",dfs(n+flag,m)+flag);
    return 0;
}