POJ 3278 Catch That Cow(BFS,板子题)
POJ 3278 Catch That Cow(BFS,板子题)
Catch That Cow
Time Limit: 2000MS | Memory Limit: 65536K | |
|---|---|---|
Total Submissions: 88732 | Accepted: 27795 |
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 17Sample Output
4Hint
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.
Source
题目链接：http://poj.org/problem?id=3278
题意：有一个农民和一头牛，他们在一个数轴上，牛在k位置保持不动，农户开始时在n位置。设农户当前在M位置，每次移动时有三种选择：1.移动到M-1；2.移动到M+1位置；3.移动到M*2的位置。问最少移动多少次可以移动到牛所在的位置。所以可以用广搜来搜索这三个状态，直到搜索到牛所在的位置。
分析：BFS的板子题，用队列做，第一次学队列，感觉自己好弱啊！现在才学，算法真的好难学，学了又怕不会用，现在不敢学太快了！
下面每一步代码我都给出了详细解释，一看就懂！
1 #include<iostream>
2 #include <algorithm>
3 #include <stdio.h>
4 #include <string.h>
5 #include<queue>
6 #define MAX 100001
7 using namespace std;
8 queue<int> q;//使用前需定义一个queue变量，且定义时已经初始化
9 bool visit[MAX];//访问空间
10 int step[MAX]; //记录步数的数组不能少
11 bool bound(int num)//定义边界函数
12 {
13 if(num<0||num>100000)
14 return true;
15 return false;
16 }
17 int BFS(int st,int end)
18 {
19 queue<int> q;//使用前需定义一个queue变量，且定义时已经初始化
20 int t,temp;
21 q.push(st);//进队列
22 visit[st]=true;
23 while(!q.empty())//重复使用时，用这个初始化
24 {
25 t=q.front();//得到队首的值
26 q.pop();//出队列,弹出队列的第一个元素，并不会返回元素的值
27 for(int i=0;i<3;++i) //三个方向搜索
28 {
29 if(i==0)
30 temp=t+1;
31 else if(i==1)
32 temp=t-1;
33 else
34 temp=t*2;
35 if(bound(temp)) //越界
36 continue;
37 if(!visit[temp])//访问空间未被标记
38 {
39 step[temp]=step[t]+1;
40 if(temp==end)
41 return step[temp];
42 visit[temp]=true;//标记此点
43 q.push(temp);//将temp元素接到队列的末端；
44 }
45 }
46 }
47 }
48 int main()
49 {
50 int st,end;
51 while(scanf("%d%d",&st,&end)!=EOF)
52 {
53 memset(visit,false,sizeof(visit));//visit数组进行初始化操作
54 if(st>=end)
55 cout<<st-end<<endl;
56 else
57 cout<<BFS(st,end)<<endl;
58 }
59 return 0;
60 }