题目链接:http://i.cnblogs.com/EditPosts.aspx?opt=1
Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7909 Accepted Submission(s):
2498
Problem 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?
* 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.今晚是大年三十除夕夜,在这里祝愿大家新年快乐!在新的一年里,开开心心,工作顺利,学习更上一层楼~n(*≧▽≦*)n
题目大意:在一个数轴上有n和k,农夫在n这个位置,奶牛在k那个位置,农夫要抓住奶牛,有两种方法:1、walking即农夫可以走x+1的位置和x-1的位置。
2、teleporting即每分钟可以走到2*x的位置。利用这两种方法,找到最快几步可以到达!
解题思路:其实就这三种情况,本来没打算用搜索的,不过发现好多东西都可以灵活运用!我们吧第一种方法看成两个方向,用dir[2]={1,-1,}表示,然后进行广搜!第二种方法自然就要特殊判断一下了!有一个地方要特殊判断一下,就是因为数据比较大,vis[ss.x]可能会超范围导致RE,所以都加一行判断使其保证在题目范围中。
详见代码。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cmath> 5 #include <queue> 6 7 using namespace std; 8 9 int dir[3]= {1,-1}; 10 11 struct node 12 { 13 int x,step; 14 } s,ss; 15 16 int bfs(int n,int k) 17 { 18 queue<node>q,qq; 19 s.x=n; 20 s.step=0; 21 int vis[100010]= {0}; 22 q.push(s); 23 while (!q.empty()) 24 { 25 s=q.front(); 26 q.pop(); 27 if (s.x==k) 28 return s.step; 29 for (int i=0; i<2; i++) 30 { 31 ss.x=s.x+dir[i]; 32 ss.step=s.step+1; 33 if (ss.x>=0&&ss.x<=100000) 34 if (!vis[ss.x]) 35 { 36 vis[ss.x]=1; 37 q.push(ss); 38 } 39 } 40 ss.x=s.x*2; 41 ss.step=s.step+1; 42 if (ss.x>=0&&ss.x<=100000) 43 { 44 if (!vis[ss.x]) 45 { 46 vis[ss.x]=1; 47 q.push(ss); 48 } 49 } 50 } 51 return 0; 52 } 53 54 int main () 55 { 56 int n,k; 57 while (~scanf("%d%d",&n,&k)) 58 { 59 printf ("%d\n",bfs(n,k)); 60 } 61 return 0; 62 }