백준 1012 c++

BFS를 이용해 몇개의 덩어리가 있는지 판단하는 문제이다.

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;

int N, M, K;
vector <vector<int>> map(50, vector<int>(50, 0));
vector <vector<bool>> visited(50, vector<bool>(50, false));

void reinit_map_and_visited()
{
	int i, j;

	for (i = 0; i < 50; i++)
	{
		for (j = 0; j < 50; j++)
		{
			//i = y축, j는 x축
			map[i][j] = 0;
			visited[i][j] = false;
		}
	}

	return;
}

void input_map()
{
	int i, X, Y;

	for (i = 0; i < K; i++)
	{
		cin >> X >> Y;//a = x, b = y
		map[X][Y] = 1;
	}

	return;
}

void BFS(int start_x, int start_y)
{
	int current_x, current_y, next_x, next_y;
	queue <pair<int, int>> q;
	vector<vector<int>> direction = { {1, 0}, {0, -1}, {-1, 0}, {0, 1}};
	int i;

	visited[start_y][start_x] = true;
	q.push({ start_x, start_y });
	while (q.empty() == false)
	{
		current_x = q.front().first;
		current_y = q.front().second;
		q.pop();
		for (i = 0; i < direction.size(); i++)
		{
			next_x = current_x + direction[i][0];
			next_y = current_y + direction[i][1];
			if ((next_x >= 0 && next_x < M) && (next_y >= 0 && next_y < N) && (map[next_y][next_x] == 1) && visited[next_y][next_x] == false)
			{
				visited[next_y][next_x] = true;
				q.push({ next_x, next_y });
			}
			else
			{
				;
			}
		}
	}

	return;
}

int find_result()
{
	int i, j, count = 0;

	for (i = 0; i < N; i++)
	{
		for (j = 0; j < M; j++)
		{
			if (visited[i][j] == false && map[i][j] == 1)
			{
				BFS(j, i);
				count++;
			}
		}
	}

	return count;
}

int main(void)
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	int T;

	cin >> T;
	while (T > 0)
	{
		cin >> N >> M >> K;
		input_map();
		cout << find_result() << "\n";
		reinit_map_and_visited();
		T--;
	}

	return 0;
}