[BJ] 2564 경비원 - JAVA

1. 문제 유형 : 구현

2. 시간복잡도 : O(N)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(in.readLine());
		int width = Integer.parseInt(st.nextToken());
		int height = Integer.parseInt(st.nextToken());
		int N = Integer.parseInt(in.readLine());
		int[][] pos = new int[N+1][3];
		for (int i = 0; i < N+1; i++) {
			st = new StringTokenizer(in.readLine());
			int dir = Integer.parseInt(st.nextToken());
			int p = Integer.parseInt(st.nextToken());
			pos[i][0] = dir;
			if(dir == 1) {
				pos[i][1] = 0; // 북
				pos[i][2] = p;
			}else if(dir == 2) {
				pos[i][1] = height; // 남
				pos[i][2] = p;
			}else if(dir == 3) {
				pos[i][1] = p; // 서
				pos[i][2] = 0;
			}else if(dir == 4) {
				pos[i][1] = p; // 동
				pos[i][2] = width;
			}
		}
		
		int answer = 0;
		for (int i = 0; i < N; i++) {
			if(pos[N][0] + pos[i][0] == 3) {
				// 남북 평행
				answer += height + Math.min(pos[N][2]+pos[i][2], 2*width - pos[N][2] - pos[i][2]);
			}else if(pos[N][0] + pos[i][0] == 7) {
				// 동서 평행
				answer += width + Math.min(pos[N][1]+pos[i][1], 2*height - pos[N][1] - pos[i][1]);
			}else {
				// 평행X
				answer += Math.abs(pos[N][1] - pos[i][1]) + Math.abs(pos[N][2] - pos[i][2]);
			}
		}
		System.out.println(answer);
	}

}