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 = 2147483647 ;
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=0 ; i<3 ; ++i ) {
                        switch ( i ) {
                                case 0 :{
                                        temp = tmp - 1 ;
                                        break;
                                }
                                case 1 :{
                                        temp = tmp + 1 ;
                                        break;
                                }
                                case 2 :{
                                        temp = tmp * 2 ;
                                        break;
                                }
                        }
                        
                        if ( temp > Max_Limit || temp < 0 ) continue ; 
                        if ( !vis[ temp ] ) {
                                Q.push( temp ) ;
                                step[ temp ] = step[ tmp ] + 1 ;
                                vis[ temp ] = true ;
                                if ( Check ( temp ) ) {
                                        Get_Target = true ; 
                                        return step[ temp ] ;
                                }
                        }
                }
        }
        if ( !Get_Target ) return -1 ; 
}

int main ( ) {
        scanf ( "%d %d" , &N , &K ) ;
        if ( N >= K ) {
                printf ( "%d
" , N - K ) ;
                goto End ;
        }
        printf ( "%d
" , BFS ( ) ) ;
        End : 
                return 0 ;
}

2016-10-19 21:59:12