POJ 3278 题解

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 78114		Accepted: 24667

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

 

 #include "iostream"
 #include "fstream"
 #include "sstream"
 #include "cstdio"
 #include "queue"

 using namespace std ;
 const int maxN = 1e5 + 1e3 ;
 const int Max_Limit = 1e5 ;
 const int INF =  ;
 typedef long long QAQ ;

 queue < int > Q ;
 bool vis[ maxN ] ;
 int step[ maxN ] ;
 int K , N ;

 bool Check ( int x_x ) { return x_x == K ? true : false ; }

 int BFS ( ) {
         int temp ;
         bool Get_Target = false ;
         Q.push ( N ) ;
         vis[ N ] = true ;
         while ( !Q.empty ( ) ) {
                 int tmp = Q.front ( ) ;
                 Q.pop ( ) ;
                 for ( int i= ; i< ; ++i ) {
                         switch ( i ) {
                                 case  :{
                                         temp = tmp -  ;
                                         break;
                                 }
                                 case  :{
                                         temp = tmp +  ;
                                         break;
                                 }
                                 case  :{
                                         temp = tmp *  ;
                                         break;
                                 }
                         }

                         if ( temp > Max_Limit || temp <  ) continue ;
                         if ( !vis[ temp ] ) {
                                 Q.push( temp ) ;
                                 step[ temp ] = step[ tmp ] +  ;
                                 vis[ temp ] = true ;
                                 if ( Check ( temp ) ) {
                                         Get_Target = true ;
                                         return step[ temp ] ;
                                 }
                         }
                 }
         }
         if ( !Get_Target ) return - ;
 }

 int main ( ) {
         scanf ( "%d %d" , &N , &K ) ;
         if ( N >= K ) {
                 printf ( "%d\n" , N - K ) ;
                 goto End ;
         }
         printf ( "%d\n" , BFS ( ) ) ;
         End :
                 return  ;
 }

2016-10-19 21:59:12