POJ3278(KB1-C 简单搜索)

Catch That Cow

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

 //2017-02-22
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <queue>

 using namespace std;

 bool book[];
 struct node
 {
     int pos, step;
 };

 int main()
 {
     int n, k;
     while(cin>>n>>k)
     {
         memset(book, , sizeof(book));
         queue<node> q;
         book[n] = ;
         node tmp;
         tmp.pos = n;
         tmp.step = ;
         q.push(tmp);
         int pos, step;
         while(!q.empty())
         {
             pos = q.front().pos;
             step = q.front().step;
             q.pop();
             if(pos == k){
                 cout<<step<<endl;
                 break;
             }
             if(pos-==k || pos+==k || *pos==k)
             {
                 cout<<step+<<endl;
                 break;
             }
             if(pos >=  && !book[pos-]){
                 tmp.pos = pos-;
                 tmp.step = step+;
                 q.push(tmp);
                 book[pos-] = ;
             }
             if(!book[pos+]){
                 book[pos+] = ;
                 tmp.pos = pos+;
                 tmp.step = step+;
                 q.push(tmp);
             }
             if(*pos<=&&!book[pos*]){
                 book[*pos] = ;
                 tmp.pos = *pos;
                 tmp.step = step+;
                 q.push(tmp);
             }
         }
     }

     return ;
 }