POJ3278 Catch That Cow 简单BFS

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本文介绍了一种使用BFS算法解决农夫约翰追捕逃牛的问题，详细解释了算法实现过程和核心步骤，包括边界检查、状态更新及路径查找，最后给出示例代码和输入输出说明。

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题解：

简单的BFS，就不说什么了，唯一注意的是要判断是不是越界了

代码

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define MAX 100000
#define LL long long
int cas=1,T;
struct Node
{
    int temp;
    int step;
};
queue<Node> que;
int vis[100000+10];
bool check(int n)
{
   if (n<0 || n>100000 )
       return true;
   else if (vis[n])
       return true;
   return false;
}
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    if (n>=k)
        printf("%d\n",n-k);
    else
    {
       Node q;
       q.temp=n;
       q.step=0;
       que.push(q);
       int flag = 0;
       while (!que.empty())
       {
          Node t = que.front();   
          que.pop();
          if (t.temp==k)
          {
              flag = 1;
              printf("%d\n",t.step);
              break;
          }

          for (int i = 0;i<3;i++)
          {
              Node a= t;
              if (i==0)
              {
                  a.temp--;
                  if (check(a.temp))
                      continue;
                  a.step++;
                  que.push(a);
                  vis[a.temp]=1;
              }
              if (i==1)
              {
                  a.temp++;
                  if (check(a.temp))
                      continue;
                  a.step++;
                  que.push(a);
                  vis[a.temp]=1;
              }
              if (i==2)
              {
                  a.temp = a.temp *2;
                  if (check(a.temp))
                      continue;
                  a.step++;
                  que.push(a);
                  vis[a.temp]=1;
              }
          }
       }
       if (!flag)
           printf("-1\n");
       memset(vis,0,sizeof(vis));

    }
    //freopen("in","r",stdin);
    //scanf("%d",&T);
    //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
    return 0;
}

题目

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit

Status

Practice

POJ 3278
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