[BaekJoon] 2583번 - 영역 구하기 [Java][C++]
[BaekJoon] 2583번 - 영역 구하기 [Java][C++]
1. 아이디어
주어진 모눈종이에서 직사각형 내부를 제외한 분리된 영역의 개수와 영역의 넓이를 오름차순으로 구하는 문제로 BFS나, DFS를 활용하면 간단하게 해결할 수 있다. 주어진 모눈종이를 시계방향으로 90도 회전했다고 생각하면 행이 N, 열이 M인 2차원 배열로 생각할 수 있으며, 직사각형이 차지하는 위치를 마킹하고 마킹되지 않은 영역에 대해 BFS, DFS로 넓이를 구하고 개수를 카운팅하면 된다.
2. 코드
1. BFS [Java]
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import java.io.*;
import java.util.*;
public class Main {
static int[] dr = {-1, 0, 1, 0};
static int[] dc = {0, 1, 0, -1};
static int n, m;
static boolean[][] grid;
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb = new StringBuilder();
StringTokenizer st = new StringTokenizer(br.readLine());
m = Integer.parseInt(st.nextToken());
n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
grid = new boolean[n][m];
while (k-- > 0) {
st = new StringTokenizer(br.readLine());
int sr = Integer.parseInt(st.nextToken());
int sc = Integer.parseInt(st.nextToken());
int er = Integer.parseInt(st.nextToken());
int ec = Integer.parseInt(st.nextToken());
for (int r = sr; r < er; r++) {
for (int c = sc; c < ec; c++) {
grid[r][c] = true;
}
}
}
int cnt = 0;
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (!grid[i][j]) {
cnt++;
list.add(bfs(i, j));
}
}
}
list.sort(Comparator.naturalOrder());
sb.append(cnt).append("\n");
for (int x : list) {
sb.append(x).append(" ");
}
System.out.println(sb);
}
static int bfs(int sr, int sc) {
Queue<int[]> q = new ArrayDeque<>();
q.offer(new int[]{sr, sc});
grid[sr][sc] = true;
int cnt = 1;
while (!q.isEmpty()) {
int[] cur = q.poll();
for (int d = 0; d < 4; d++) {
int nr = cur[0] + dr[d];
int nc = cur[1] + dc[d];
if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;
if (grid[nr][nc]) continue;
q.offer(new int[]{nr, nc});
grid[nr][nc] = true;
cnt++;
}
}
return cnt;
}
}
2. BFS [C++]
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#include <bits/stdc++.h>
using namespace std;
int dr[4] = {-1, 0, 1, 0};
int dc[4] = {0, 1, 0, -1};
int n, m;
bool grid[100][100];
int bfs(int sr, int sc) {
queue<pair<int, int>> q;
q.push({sr, sc});
grid[sr][sc] = true;
int cnt = 1;
while (!q.empty()) {
auto [r, c] = q.front();
q.pop();
for (int d = 0; d < 4; d++) {
int nr = r + dr[d];
int nc = c + dc[d];
if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;
if (grid[nr][nc]) continue;
q.push({nr, nc});
grid[nr][nc] = true;
cnt++;
}
}
return cnt;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int k;
cin >> m >> n >> k;
while (k--) {
int sr, sc, er, ec;
cin >> sr >> sc >> er >> ec;
for (int r = sr; r < er; r++) {
for (int c = sc; c < ec; c++) {
grid[r][c] = true;
}
}
}
int cnt = 0;
vector<int> v;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (!grid[i][j]) {
cnt++;
v.push_back(bfs(i, j));
}
}
}
sort(v.begin(), v.end());
cout << cnt << '\n';
for (int x : v) {
cout << x << ' ';
}
}
3. DFS [Java]
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import java.io.*;
import java.util.*;
public class Main {
static int[] dr = {-1, 0, 1, 0};
static int[] dc = {0, 1, 0, -1};
static int n, m;
static boolean[][] grid;
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb = new StringBuilder();
StringTokenizer st = new StringTokenizer(br.readLine());
m = Integer.parseInt(st.nextToken());
n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
grid = new boolean[n][m];
while (k-- > 0) {
st = new StringTokenizer(br.readLine());
int sr = Integer.parseInt(st.nextToken());
int sc = Integer.parseInt(st.nextToken());
int er = Integer.parseInt(st.nextToken());
int ec = Integer.parseInt(st.nextToken());
for (int r = sr; r < er; r++) {
for (int c = sc; c < ec; c++) {
grid[r][c] = true;
}
}
}
int cnt = 0;
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (!grid[i][j]) {
cnt++;
list.add(dfs(i, j));
}
}
}
list.sort(Comparator.naturalOrder());
sb.append(cnt).append("\n");
for (int x : list) {
sb.append(x).append(" ");
}
System.out.println(sb);
}
static int dfs(int r, int c) {
grid[r][c] = true;
int cnt = 1;
for (int d = 0; d < 4; d++) {
int nr = r + dr[d];
int nc = c + dc[d];
if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;
if (grid[nr][nc]) continue;
cnt += dfs(nr, nc);
}
return cnt;
}
}
4. DFS [C++]
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#include <bits/stdc++.h>
using namespace std;
int dr[4] = {-1, 0, 1, 0};
int dc[4] = {0, 1, 0, -1};
int n, m;
bool grid[100][100];
int dfs(int r, int c) {
grid[r][c] = true;
int cnt = 1;
for (int d = 0; d < 4; d++) {
int nr = r + dr[d];
int nc = c + dc[d];
if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;
if (grid[nr][nc]) continue;
cnt += dfs(nr, nc);
}
return cnt;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int k;
cin >> m >> n >> k;
while (k--) {
int sr, sc, er, ec;
cin >> sr >> sc >> er >> ec;
for (int r = sr; r < er; r++) {
for (int c = sc; c < ec; c++) {
grid[r][c] = true;
}
}
}
int cnt = 0;
vector<int> v;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (!grid[i][j]) {
cnt++;
v.push_back(dfs(i, j));
}
}
}
sort(v.begin(), v.end());
cout << cnt << '\n';
for (int x : v) {
cout << x << ' ';
}
}
3. 디버깅
없음.
4. 참고
없음.
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