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.
View Code
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<string.h>
4 #include<math.h>
5 const int M = 1000005;
6 int n, m, q[2*M], front, last, h[2*M];
7 int judge( int x )
8 {
9 if( x < 0 || x >= 2*M || h[x] )
10 return 0;
11 return 1;
12 }
13
14 int BFS( int a, int b )
15 {
16 front = last = 0;
17 q[front++] = a;
18 if( a == b )
19 return 0;
20 while( last < front )
21 {
22
23 int x = q[last++];
24 if( x+1 == b || x-1 == b|| 2*x == b )
25 return h[x]+1;
26 if( judge( x+1) )
27 {
28 h[x+1] = h[x]+1,q[front++] = x+1;
29 }
30 if( judge(x-1) )
31 {
32 h[x-1] = h[x]+1,q[front++] = x-1;
33 }
34 if( judge(2*x) )
35 {
36 h[2*x] = h[x]+1,q[front++] = 2*x;
37 }
38 }
39 return -1;
40 }
41
42 int main()
43 {
44 while( scanf( "%d%d", &n, &m) == 2 )
45 {
46 for( int i = 0; i < 2*M; i++ )
47 h[i] = 0;
48 printf( "%d\n", BFS( n, m ) );
49 }
50 return 0;
51 }
