poj 3278 Catch That Cow(bfs广搜)

最新推荐文章于 2024-07-24 11:36:30 发布

原创最新推荐文章于 2024-07-24 11:36:30 发布 · 348 阅读

0 ·

CC 4.0 BY-SA版权

搜索专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一个有趣的算法问题——农夫如何用最短时间找到他的牛。通过三种移动方式：步进加1、步进减1及坐标翻倍，利用广度优先搜索（BFS）策略快速定位目标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 49724		Accepted: 15590

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.

题意：农夫想快速找到自己的牛，在一维坐标中，农夫坐标为N，牛的坐标为K，农夫有3种方式移动：

1.使自己坐标加1

2.使自己坐标减1

3.使自己坐标乘2倍

（如：若农夫在4，下一秒可到达3,5,8）

求农夫找到牛的最短时间。

题解：将农夫可做的3种状态成为当前的子树，进行广搜，到达牛的坐标K立即结束输出答案。

（第二次使用bfs~居然1A了，可见这是道水题~）

AC代码：

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
int n,k;
int a[100005];
queue<int> q;
int bfs(){
    q.push(n);
    a[n]=0;
    while(!q.empty()){
        int x;
        x=q.front();
        if(x==k)
            return printf("%d\n",a[k]);
        q.pop();
        if(x+1<=100000&&a[x+1]==-1)
        {q.push(x+1);a[x+1]=a[x]+1; }
        if(x-1>=0&&a[x-1]==-1)
        {q.push(x-1);a[x-1]=a[x]+1; }
        if(2*x<=100000&&a[2*x]==-1)
        {q.push(2*x);a[2*x]=a[x]+1; }
    }
}

int main(){
    while(cin>>n>>k){
        while(!q.empty()){
            q.pop();
        }
        for(int i=0;i<100005;i++)
            a[i]=-1;
        bfs();
    }
    return 0;
}