Problem Solving/Baekjoon
[백준] 3085 사탕 게임 - Brute Force / Java
graycode
2022. 5. 17. 23:55
• 문제 링크
3085번: 사탕 게임
예제 3의 경우 4번 행의 Y와 C를 바꾸면 사탕 네 개를 먹을 수 있다.
www.acmicpc.net
• 풀이 과정
• 풀이 코드
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
public class Main {
public static int n;
public static char[][] box;
public static int max = 0;
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
n = Integer.parseInt(br.readLine());
box = new char[n][n];
for (int i = 0; i < n; i++) {
String line = br.readLine();
box[i] = line.toCharArray();
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - 1; j++) {
char swap = box[i][j];
box[i][j] = box[i][j + 1];
box[i][j + 1] = swap;
search();
swap = box[i][j];
box[i][j] = box[i][j + 1];
box[i][j + 1] = swap;
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - 1; j++) {
char swap = box[j][i];
box[j][i] = box[j + 1][i];
box[j + 1][i] = swap;
max = Math.max(search(), max);
swap = box[j][i];
box[j][i] = box[j + 1][i];
box[j + 1][i] = swap;
}
}
bw.write(max + "\n");
bw.flush();
}
private static int search() {
for (int i = 0; i < n; i++) {
int cnt = 1;
for (int j = 0; j < n - 1; j++) {
if (box[i][j] == box[i][j + 1]) {
cnt++;
} else {
cnt = 1;
}
max = Math.max(max, cnt);
}
}
for (int i = 0; i < n; i++) {
int cnt = 1;
for (int j = 0; j < n - 1; j++) {
if (box[j][i] == box[j + 1][i]) {
cnt++;
} else {
cnt = 1;
}
max = Math.max(max, cnt);
}
}
return max;
}
}