Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12798 Accepted Submission(s): 3950
Problem 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 USACO 2007 Open Silver
解析：简单BFS。易知农夫走的范围不会超过2*k，超过2*k就得不到最小值。
``` #include
const int MAXN = 100000+5;
int dis[2MAXN];
int q[2MAXN];
int bfs(int n, int k)
{
memset(dis, -1, sizeof dis);
dis[n] = 0;
int l = 0, r = 0;
q[r++] = n;
while(l < r){
int h = q[l++];
if(h == k)
return dis[k];
int tmp = h-1;
if(tmp >= 0 && dis[tmp] == -1){
q[r++] = tmp;
dis[tmp] = dis[h]+1;
}
tmp = h+1;
if(tmp <= 2k && dis[tmp] == -1){
q[r++] = tmp;
dis[tmp] = dis[h]+1;
}
tmp = h2;
if(tmp <= 2*k && dis[tmp] == -1){
q[r++] = tmp;
dis[tmp] = dis[h]+1;
}
}
}
int main()
{
int n, k;
while(~scanf("%d%d", &n, &k)){
int res = bfs(n, k);
printf("%d
", res);
}
return 0;
}