【OpenJ_Bailian - 4001】 Catch That Cow（bfs+优先队列）

Catch That Cow

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Descriptions:

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.

题目链接：

https://vjudge.net/problem/OpenJ_Bailian-4001

 

题目大意
农夫约翰获知了一头逃走的牛的位置，想要将它立即抓住。他从数轴上的一个点 N (0 ≤ N ≤ 100,000) 出发，牛位于相同数轴上的点 K (0 ≤ K ≤ 100,000)。农夫约翰有两种前进方式：行走和神行。

行走：农夫约翰可以从任何 X 移动到点 X - 1 或 X + 1，只需要一分钟；
神行：农夫约翰可以从任何点 X 移动到点 2 × X，也只需要一分钟。
如果那头逃走的牛并不知道对它实施的抓捕行动，因此完全不移动，那么农夫约翰花多少时间可以抓住这头牛？
Input
第一行包含两个以空格分隔的整数： N 和 K
Output
第一行输出农夫约翰抓住逃走的牛，所花费的最短时间 (分钟)。
Hint
农夫约翰抓住逃走的牛的最快方式，是按如下路径移动：5-10-9-18-17，这花费了 4 分钟。

 

这题我的想法就是bfs，加个优先队列吧，毕竟求最小话费时间，优先队列是一定没有错的，个人习惯写法。大致判断一下农夫下一步该走到哪里，那个地方是否满足2个条件：是否在0-100000之间；这个点是否走过。符合这两点就能入队了。break的判断就是，农夫现在的地方等于牛的地方，写的比较粗糙，见谅。

 

AC代码

 #include <iostream>
 #include <cstdio>
 #include <fstream>
 #include <algorithm>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <string>
 #include <cstring>
 #include <map>
 #include <stack>
 #include <set>
 using namespace std;
 int N,K;
 int vis[];//路径判断是否走过
 struct node
 {
     int n,k;//n为现在的地方，k为牛的地方
     int step;//步数即时间
     friend bool operator<(node a,node b)//步数小的现出来，即时间少
     {
         return a.step>b.step;
     }
 };
 priority_queue<node>q;
 void bfs()
 {
     node now;//初始化now
     now.n=N;
     now.k=K;
     now.step=;
     q.push(now);
     while(!q.empty())
     {
         node mid=q.top();
         q.pop();
         if(mid.n==mid.k)//农夫现在的地方等于牛的地方，即找到牛了
         {
             cout << mid.step <<endl;
             return;
         }
         //农夫的三种走法判断是否符合题意
         if(mid.n+>=&&mid.n+<=&&!vis[mid.n+])
         {
             node last;//更新队列
             last.k=K;
             last.n=mid.n+;
             last.step=mid.step+;//步数+1
             vis[mid.n+]=;//标记路径
             q.push(last);//入队
         }
         if(mid.n->=&&mid.n-<=&&!vis[mid.n-])
         {
             node last;
             last.k=K;
             last.n=mid.n-;
             last.step=mid.step+;
             vis[mid.n-]=;
             q.push(last);
         }
         if(mid.n*>=&&mid.n*<=&&!vis[mid.n*])
         {
             node last;
             last.k=K;
             last.n=mid.n*;
             last.step=mid.step+;
             vis[mid.n*]=;
             q.push(last);
         }
     }
 }
 int main()
 {
     memset(vis,,sizeof(vis));//初始化路径都为0，即没有走过
     cin >> N >> K;
     vis[N]=;
     bfs();
     return ;
 }