POJ 3278 Catch That Cow

Catch That Cow

Time Limit:2000 MS		Memory Limit:65536 K
Total Submissions:44070		Accepted:13764

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN(0 ≤N≤ 100,000) on a number line and the cow is at a pointK(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 pointXTo the pointsX-1 orX+ 1 in a single minute
* Teleporting: FJ can move from any pointXTo the point 2 ×XIn 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: NAnd 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

The simple BFS application officially started with POJ, and the start of the poj plan meant that my life would not be as easy as before.

Let me give myself a sentence: I want to have fun with ACM. I have eight words: "Sea of questions and tactics! Question sea tactics !"

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include #include 
     
      const int N =  10000010;using namespace std;int n,k;struct node{    int wz;    int ans;}q[N];int vis[N];int jz[] = {1,-1};void BFS(){    int s = 0,e = 0;    node f,t;    f.ans = 0;    f.wz = n;    q[e++] = f;    vis[n] = 1;    while(s
      
       =0 && f.wz <=100000) { f.ans = t.ans + 1; vis[f.wz] = 1; q[e++] = f; } } }}int main(){ while(~scanf("%d%d",&n,&k)) { memset(vis,0,sizeof(vis)); BFS(); } return 0;}