poj 3278--Catch That Cow
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.
初学bfs最好的题了，比较简单，当初刚学的时候就做了好几遍这个--
View Code
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #include <algorithm> 5 #include <cstdlib> 6 using namespace std; 7 const int N=100010; 8 struct num 9 { 10 int flag; 11 int step; 12 }p[N]; 13 int main() 14 { 15 int n,k; 16 queue<int>q; 17 memset(p,0,sizeof(p)); 18 scanf("%d%d",&n,&k); 19 q.push(n); 20 p[n].flag=1; 21 while(!q.empty()) 22 { 23 n=q.front(); 24 if(n==k) 25 break; 26 q.pop(); 27 if((n-1)>=0&&(n-1)<=100000&&p[n-1].flag==0) 28 { 29 q.push(n-1); 30 p[n-1].flag=1; 31 p[n-1].step=p[n].step+1; 32 /*if(n-1==k) 33 { 34 printf("%d\n",p[n-1].step); 35 break; 36 }*/ 37 } 38 if((n+1)>=0&&(n+1)<=100000&&p[n+1].flag==0) 39 { 40 q.push(n+1); 41 p[n+1].flag=1; 42 p[n+1].step=p[n].step+1; 43 /*if(n+1==k) 44 { 45 printf("%d\n",p[n+1].step); 46 break; 47 }*/ 48 } 49 if((n*2)>=0&&(n*2)<=100000&&p[n*2].flag==0) 50 { 51 q.push(n*2); 52 p[n*2].flag=1; 53 p[n*2].step=p[n].step+1; 54 /*if(n*2==k) 55 { 56 printf("%d\n",p[n*2].step); 57 break; 58 }*/ 59 } 60 } 61 printf("%d\n",p[k].step); 62 return 0; 63 }
