Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 169075 | Accepted: 51762 |
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.
Source
首先拿道这个题以为是数论，用二进制捣鼓了半天依然没有什么发现
想到搜索，DFS or BFS？
DFS一般要整个遍历一遍再求解
BFS更适合解决一些最值问题，在最坏情况下和DFS的复杂度一致，但是一般情况下最值按照BFS都更易求
所以遇到一些最值搜索问题时可以考虑用BFS
变形的最短路应该看出来
无论BFS还是DFS都一定要注意标记已经走过的点！！！！！
这题在一些细节上还有待注意，比如说起止点可以是0，还有可以先乘到终止点后面再减回来
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <queue> 5 #include <stack> 6 #include <vector> 7 #include <iostream> 8 #include "algorithm" 9 using namespace std; 10 const int MAX=100005; 11 int n,m,ans; 12 bool co[MAX]; 13 struct num{ 14 int x,y; 15 }; 16 queue <num> q; 17 int main(){ 18 freopen ("cow.in","r",stdin); 19 freopen ("cow.out","w",stdout); 20 int i,j; 21 scanf("%d%d",&n,&m); 22 memset(co,true,sizeof(co)); 23 num a,b;a.x=n,a.y=0; 24 q.push(a); 25 co[a.x]=false; 26 while (!q.empty()){ 27 a=q.front();q.pop(); 28 if (a.x*2<MAX && co[a.x*2]){ 29 b.x=a.x*2,b.y=a.y+1; 30 q.push(b); 31 co[b.x]=false; 32 if (b.x==m){ans=b.y;break;} 33 } 34 if (a.x+1<MAX && co[a.x+1]){ 35 b.x=a.x+1,b.y=a.y+1; 36 q.push(b); 37 co[b.x]=false; 38 if (b.x==m){ans=b.y;break;} 39 } 40 if (a.x-1>=0 && co[a.x-1]){ 41 b.x=a.x-1,b.y=a.y+1; 42 q.push(b); 43 co[b.x]=false; 44 if (b.x==m){ans=b.y;break;} 45 } 46 } 47 printf("%d",ans); 48 return 0; 49 }