유기농 배추 (feat. 백준, 1012번)

유기농 배추
https://www.acmicpc.net/problem/1012

1012번: 유기농 배추

 

반응형

 

// woohyeon.kim
// kim519620.tistory.com
#include <iostream>
using namespace std;
int T, M, N, K, arr[50][50], dx[4], dy[4], ans;
bool visit[50][50];
bool dfs(register int x, register int y) {
	if (visit[x][y])		return false;
	visit[x][y] = true;
	for (register int d = 0, nx, ny; d < 4; ++d) {
		nx = x + dx[d], ny = y + dy[d];
		if (0 <= nx && nx < N && 0 <= ny && ny < M && !visit[nx][ny] && arr[nx][ny] == 1)
			dfs(nx, ny);
	}
	return true;
}
int main() {
	ios::sync_with_stdio(false), cin.tie(NULL);
	dx[0] = 1, dx[1] = -1, dx[2] = dx[3] = 0;
	dy[0] = dy[1] = 0, dy[2] = 1, dy[3] = -1;
	cin >> T;
	for (register int t = 0; t < T; ++t) {
		cin >> M >> N >> K;
		for (register int n = 0, m; n < N; ++n)
			for (m = 0; m < M; ++m)
				arr[n][m] = 0, visit[n][m] = false;
		for (register int k = 0, x, y; k < K; ++k)
			cin >> x >> y, arr[y][x] = 1;
		ans = 0;
		for (register int n = 0, m; n < N; ++n)
			for (m = 0; m < M; ++m)
				if (arr[n][m] && dfs(n, m))
					++ans;
		cout << ans << "\n";
	}
	return 0;
}
// *&)*@*

DFS 풀이 방법

 

// woohyeon.kim
// kim519620.tistory.com
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
int T, M, N, K, arr[50][50], dx[4], dy[4], ans;
bool visit[50][50];
bool bfs(register int x, register int y) {
	if (visit[x][y])		return false;
	visit[x][y] = true;
	queue<pair<int, int>> q;
	q.push(make_pair(x, y));
	while (!q.empty()) {
		auto v = q.front(); q.pop();
		x = v.first, y = v.second;
		for (register int d = 0, nx, ny; d < 4; ++d) {
			nx = x + dx[d], ny = y + dy[d];
			if (0 <= nx && nx < N && 0 <= ny && ny < M && !visit[nx][ny] && arr[nx][ny] == 1)
				visit[nx][ny] = true, q.push(make_pair(nx, ny));
		}
	}
	return true;
}
int main() {
	ios::sync_with_stdio(false), cin.tie(NULL);
	dx[0] = 1, dx[1] = -1, dx[2] = dx[3] = 0;
	dy[0] = dy[1] = 0, dy[2] = 1, dy[3] = -1;
	cin >> T;
	for (register int t = 0; t < T; ++t) {
		cin >> M >> N >> K;
		for (register int n = 0, m; n < N; ++n)
			for (m = 0; m < M; ++m)
				arr[n][m] = 0, visit[n][m] = false;
		for (register int k = 0, x, y; k < K; ++k)
			cin >> x >> y, arr[y][x] = 1;
		ans = 0;
		for (register int n = 0, m; n < N; ++n)
			for (m = 0; m < M; ++m)
				if (arr[n][m] && bfs(n, m))
					++ans;
		cout << ans << "\n";
	}
	return 0;
}
// *&)*@*

BFS 풀이 방법

 

연결된 집합이 총 몇개인지를 구하는 문제이므로 DFS, BFS 모두 풀이가 가능합니다.