Catch That Cow——广搜

「爱情、让人受尽委屈。」 2022-06-06 08:58 145阅读 0赞

# Catch That Cow #

## Problem 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.\[/hint\]

题目大意：农夫的牛丢了，现在给农夫的位置N，和牛的位置K。一维坐标。现在假如牛不动，农夫去找牛，有两种走路方式，一种x-1，和x+1用时1。

一种2\*x，用时1 。求最短时间。

简单广搜，搜索入门题。

这道题要标记找过的不能找了，超出范围的不能找了，否则会时间超限。

#include<bits/stdc++.h>///Catch That Cow
    using namespace std;
    int to[2]={1,-1};
    int a,b,sum;
    int vis[100000];
    struct place
    {
        int x,time;
    };
    int check(place k)
    {
        if(k.x<0||k.x>100000||vis[k.x]==1)
            return 0;
        return 1;
    }
    int bfs(place n)
    {
        place m,next;
        queue<place>w;
        w.push(n);
        while(!w.empty())
        {
            m=w.front();
            w.pop();
            if(m.x==b)
                return m.time;
            for(int i=0;i<2;i++)
            {
                next.x=m.x+to[i];
                next.time=m.time+1;
                if(next.x==b)
                    return next.time;
                if(check(next))
                {
                    w.push(next);
                    vis[next.x]=1;
                }
            }
            next.x=m.x*2;
            next.time=m.time+1;
            if(next.x==b)
                return next.time;
            if(check(next))
            {
                w.push(next);
                vis[next.x]=1;
            }
        }
        return 0;
    }
    int main()
    {
        int i,j,t;
        place x1;
        while(~scanf("%d %d",&a,&b))
        {
            memset(vis,0,sizeof(vis));
            x1.x=a;
            x1.time=0;
            vis[x1.x]=1;
            sum=0;
            sum=bfs(x1);
            printf("%d\n",sum);
        }
        return 0;
    }