2589 보물섬

https://www.acmicpc.net/problem/2589

이 문제는 BFS 문제로 쉽게 풀 수 있습니다.

BFS가 실행될 때마다 각 최대값을 검색하고 가장 가까운 값과 비교하면서 가장 큰 값을 찾습니다.

package BFS;

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 N2589 {
    public static class Pos{
        int r,c,dist;
        public Pos(int r, int c, int dist){
            this.r = r;
            this.c = c;
            this.dist = dist;
        }
    }

    public static void main(String() args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int row = Integer.parseInt(st.nextToken());
        int col = Integer.parseInt(st.nextToken());
        char()() world = new char(row)(col);
        Queue<Pos> q  = new LinkedList<>();
        for (int i = 0; i < row; i++) {
            String a = br.readLine();
            for (int j = 0; j < col; j++) {
                world(i)(j) = a.charAt(j);
            }
        }

        int maximum = 0;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if(world(i)(j) == 'L'){
                    boolean()() visited = new boolean(row)(col);
                    q.add(new Pos(i,j,0));
                    visited(i)(j) = true;
                    int result = Bfs(world,visited,row,col,q);
                    if(maximum < result){
                        maximum = result;
                    }
                }
            }
        }
        System.out.println(maximum);


    }

    public static int Bfs(char()() world, boolean()() visited, int row, int col,Queue<Pos> q){
        int max = 0;
        int() DR = {1,0,-1,0};
        int() DC = {0,1,0,-1};
        while(!
q.isEmpty()){
            Pos Cur = q.poll();
            if(Cur.dist>max){
                max = Cur.dist;
            }
            for (int i = 0; i < DR.length; i++) {
                int nextR = Cur.r + DR(i);
                int nextC = Cur.c + DC(i);

                if(nextR < 0 || nextC < 0 || nextR >= row || nextC >= col) continue;

                if(world(nextR)(nextC) == 'L' && !
visited(nextR)(nextC)){
                    visited(nextR)(nextC) = true;
                    q.add(new Pos(nextR,nextC,Cur.dist+1));
                }
            }
        }
        return max;
    }
}