[BaekJoon] 1926번 - 그림 [Java][C++]

문제 링크

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1. 아이디어

도화지에 있는 그림의 수와 가장 넓은 그림의 넓이를 구하는 문제로 도화지의 각 칸에 대해 그림의 일부이면 BFS나, DFS를 통해 넓이를 구하고 방문 체크를 하면 간단하게 해결할 수 있다.

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2. 코드

1. BFS [Java]

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 int[][] grid;

    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());
        m = Integer.parseInt(st.nextToken());

        grid = new int[n][m];
        for (int i = 0; i < n; i++) {
            st = new StringTokenizer(br.readLine());
            for (int j = 0; j < m; j++) {
                grid[i][j] = Integer.parseInt(st.nextToken());
            }
        }

        int cnt = 0;
        int max = 0;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) {
                    cnt++;
                    max = Math.max(max, bfs(i, j));
                }
            }
        }

        System.out.println(cnt);
        System.out.println(max);
    }

    static int bfs(int sr, int sc) {
        Queue<int[]> q = new ArrayDeque<>();
        q.offer(new int[]{sr, sc});

        grid[sr][sc] = 0;

        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] == 0) continue;

                q.offer(new int[]{nr, nc});
                grid[nr][nc] = 0;
                cnt++;
            }
        }

        return cnt;
    }
}

2. BFS [C++]

#include <bits/stdc++.h>
using namespace std;

int dr[4] = {-1, 0, 1, 0};
int dc[4] = {0, 1, 0, -1};
int n, m;
int grid[500][500];

int bfs(int sr, int sc) {
    queue<pair<int, int>> q;
    q.push({sr, sc});

    grid[sr][sc] = 0;

    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] == 0) continue;

            q.push({nr, nc});
            grid[nr][nc] = 0;
            cnt++;
        }
    }

    return cnt;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    cin >> n >> m;

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> grid[i][j];
        }
    }

    int cnt = 0;
    int mx = 0;

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (grid[i][j] == 1) {
                cnt++;
                mx = max(mx, bfs(i, j));
            }
        }
    }

    cout << cnt << '\n';
    cout << mx << '\n';
}

3. DFS [Java]

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 int[][] grid;

    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());
        m = Integer.parseInt(st.nextToken());

        grid = new int[n][m];
        for (int i = 0; i < n; i++) {
            st = new StringTokenizer(br.readLine());
            for (int j = 0; j < m; j++) {
                grid[i][j] = Integer.parseInt(st.nextToken());
            }
        }

        int cnt = 0;
        int max = 0;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) {
                    cnt++;
                    max = Math.max(max, dfs(i, j));
                }
            }
        }

        System.out.println(cnt);
        System.out.println(max);
    }

    static int dfs(int r, int c) {
        grid[r][c] = 0;
        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] == 0) continue;

            cnt += dfs(nr, nc);
        }

        return cnt;
    }
}

4. DFS [C++]

#include <bits/stdc++.h>
using namespace std;

int dr[4] = {-1, 0, 1, 0};
int dc[4] = {0, 1, 0, -1};
int n, m;
int grid[500][500];

int dfs(int r, int c) {
    grid[r][c] = 0;
    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] == 0) continue;

        cnt += dfs(nr, nc);
    }

    return cnt;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    cin >> n >> m;

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> grid[i][j];
        }
    }

    int cnt = 0;
    int mx = 0;

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (grid[i][j] == 1) {
                cnt++;
                mx = max(mx, dfs(i, j));
            }
        }
    }

    cout << cnt << '\n';
    cout << mx << '\n';
}

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3. 디버깅

없음.

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4. 참고

없음.

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