POJ - 3278-Catch That Cow(BFS)

最新推荐文章于 2024-07-24 11:36:30 发布

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最新推荐文章于 2024-07-24 11:36:30 发布

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分类专栏： ACM搜索文章标签： bfs poj

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本文介绍了一个名为CatchThatCow的问题，讲述了农民约翰如何利用步行和瞬移两种方式在一维数轴上追捕逃跑的奶牛。文章通过示例详细解释了问题背景、输入输出格式，并给出了一种使用广度优先搜索的解决方案。

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Catch That Cow

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 101593 Accepted: 31754

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.

题解：n 到k 有三种走法 n-1,n+1, 2*n
特殊情况：若n=k 返回0
若n>k 只能走n-1

#include<stdio.h>
#include<iostream>
#include<queue>
using namespace std;
const int maxn=100000;
struct point
{
    int pos;
    int step;
}now,next;
 int n,k;
queue<point> Q;
bool vis[maxn];
int bfs()
{
    while(!Q.empty()) Q.pop();

    now.pos=n;
    now.step=0;
    vis[now.pos]=true;
    Q.push(now);
    while(!Q.empty())
    {

        now=Q.front();
        Q.pop();
        next=now;
        for(int i=0;i<3;i++)
        {

            if(i==0) next.pos=now.pos+1;
            if(i==1) next.pos=now.pos-1;
            if(i==2) next.pos=now.pos*2;
            next.step=now.step+1;

            if(next.pos==k) return next.step;
            if(next.pos<0||next.pos>maxn) continue;
            if(!vis[next.pos])
            {

                vis[next.pos]=true;
                Q.push(next);
            }
        }
    }
}
int main()
{
     while(~(scanf("%d%d",&n,&k)))
     {
          if(n<k)
            cout<<bfs()<<endl;
          else if(n==k)
            cout<<0<<endl;
          else
            cout<<n-k<<endl;
     }
    return 0;
}