HUD 2717 Catch That Cow

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
#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\n", res);

}

return 0;

}