POJ 3278 Catch That Cow（BFS）

最新推荐文章于 2024-09-04 11:33:10 发布

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最新推荐文章于 2024-09-04 11:33:10 发布

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本文链接：https://blog.youkuaiyun.com/qq_32473657/article/details/51981645

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本文介绍了一个经典的搜索问题：农民约翰如何用最短时间找到一只静止不动的逃跑的奶牛。通过使用广度优先搜索算法（BFS），文章详细解释了如何从起点到达目标点所需的最少步数，包括步行和瞬间传送两种移动方式。

不要小看任何一道题。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int man=1e7+5;
int a[man];
int book[man];
int step[man];
int n,k;
int judge(int s)
{
    if(s>=0&&s<=100000&&book[s]==0)
        return 1;
    return 0;
}
int bfs(int n,int k)
{
    memset(book,0,sizeof(book));
    queue<int>Q;
    Q.push(n);
    book[n]=1;
    step[n]=0;
    while(!Q.empty())
    {
        int b=Q.front();
        Q.pop();
        if(b==k)
        {
            return step[k];
        }
        //for(int i=0; i<3; i++)
        {
            //if(i==0)

                int a=b-1;
                if(judge(a))
                    Q.push(a),book[a]=1,step[a]=step[b]+1;//,printf("%d\n",a);

            //if(i==1)

                a=b+1;
                if(judge(a))
                    Q.push(a),book[a]=1,step[a]=step[b]+1;//,printf("%d\n",a);

            //if(i==2)

                a=b*2;
                if(judge(a))        //;<span style="white-space:pre">				</span>//在这SB了好多次
                Q.push(a),book[a]=1,step[a]=step[b]+1;//,printf("%d\n",a);

        }
    }
    return 0;<span style="white-space:pre">						</span>//细节 必须有 
}
int main()
{
    //while(~
          scanf("%d%d",&n,&k);//)
    {
        int an=bfs(n,k);
        printf("%d\n",an);
    }
}

Catch That Cow

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status Practice POJ 3278

Appoint description:

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