백준 1697. 숨바꼭질 :: 돼지개발자

출저 : https://www.acmicpc.net/problem/1697

"단순 BFS"

수빈이가 이동할 수 있는 방법을 길이 3인 배열로 만들어 처리했음.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;
 
public class Main {
 
    static int N,K;
    static int INF = 100001;
    static int[] visited = new int[INF];
    static Queue<Integer> q = new LinkedList<>();
    
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        
        N = Integer.parseInt(st.nextToken());
        K = Integer.parseInt(st.nextToken());
        
        q.add(N);
        
        while(!q.isEmpty()) {
            int cur = q.poll();
            
            if(cur == K) {
                break ;
            }
            
            int[] nexts = { cur+1, cur-1, 2*cur};
            
            for (int i = 0; i < 3; i++) {
                int next = nexts[i];
                if(next < 0 || INF <= next || visited[next] != 0) continue;
                
                visited[next] = visited[cur]+1;
                q.add(next);
            }    
        }
        System.out.println(visited[K]);
    }
}