Catch That Cow
| 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
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.
题意：n通过几次变化能到k,n可以加一，减一，乘二；
bfs搜索一遍即可
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 }