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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 64176 | Accepted: 20156 |
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
Output
Sample Input
5 17
Sample Output
4
Hint
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 #include <string.h> 6 #include <algorithm> 7 using namespace std; 8 int n,k,num[100000 + 5]; 9 10 void bfs(int y) 11 { 12 queue<int> q; 13 q.push(y); 14 while(q.size()) 15 { 16 int fx; 17 int x = q.front(); 18 q.pop(); 19 if(x == k) 20 break; 21 for(int i = 0; i < 3; i++) 22 { 23 if(i == 0) 24 fx = x + 1; 25 else if(i == 1) 26 fx = x - 1; 27 else if(i == 2) 28 fx = x * 2; 29 if(fx >= 0 && fx <= 100000 && num[fx] > num[x]) 30 { 31 num[fx] = num[x] + 1; 32 q.push(fx); 33 } 34 } 35 } 36 } 37 int main() 38 { 39 while(scanf("%d%d", &n,&k) != EOF) 40 { 41 for(int i = 0; i < 100005; i++) 42 num[i] = 100000000; 43 num[n] = 0; 44 bfs(n); 45 printf("%d\n",num[k]); 46 } 47 48 return 0; 49 }
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原文地址：http://www.cnblogs.com/zhaopAC/p/4969546.html
