백준 문제풀이

백준 10026번 - C++

diligent_gideok 2022. 4. 30. 06:38
#include <bits/stdc++.h>
using namespace std;
#define X first
#define Y second
char board[101][101];
bool vis[101][101];


int dx[4] = { 1,0,-1,0 };
int dy[4] = { 0,1,0,-1 };

int main(void) {
	ios::sync_with_stdio(0);
	cin.tie(0);
	queue<pair<int, int> > Q;
	int n, val1 = 0,val2=0;
	cin >> n;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			cin >> board[i][j];
		}
	}

	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			if (vis[i][j] == 0) {
				val1++;
				vis[i][j] = 1;
				Q.push({ i,j });
				char c = board[i][j];
				while (!Q.empty()) {
					pair<int, int> cur = Q.front(); Q.pop();
					//cout << '(' << cur.X << ", " << cur.Y << ") -> ";
					for (int dir = 0; dir < 4; dir++) {
						int nx = cur.X + dx[dir];
						int ny = cur.Y + dy[dir];
						if (nx < 0 || nx >= n || ny < 0 || ny >= n) continue;
						if (vis[nx][ny] || board[nx][ny] != c) continue;
						vis[nx][ny] = 1;
						Q.push({ nx,ny });
					}
				}
			}
		}
	}
	cout << val1 << ' ';
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			if(board[i][j]=='R') board[i][j]='G';
		}
	}
	// 적록색약인 사람을 구하기위한 방문배열 초기화
	for (int i = 0; i < n; i++)
		fill(vis[i], vis[i] + n, false);

	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			if (vis[i][j] == 0) {
				val2++;
				vis[i][j] = 1;
				Q.push({ i,j });
				char c = board[i][j];
				while (!Q.empty()) {
					pair<int, int> cur = Q.front(); Q.pop();
					//cout << '(' << cur.X << ", " << cur.Y << ") -> ";
					for (int dir = 0; dir < 4; dir++) {
						int nx = cur.X + dx[dir];
						int ny = cur.Y + dy[dir];
						if (nx < 0 || nx >= n || ny < 0 || ny >= n) continue;
						if (vis[nx][ny] || board[nx][ny] != c) continue;
						vis[nx][ny] = 1;
						Q.push({ nx,ny });
					}
				}
			}
		}
	}
	cout << val2;
}

'백준 문제풀이' 카테고리의 다른 글

백준 1780번 - C++  (0) 2022.04.30
백준 1074번 - C++  (0) 2022.04.30
백준 7562번 - C++  (0) 2022.04.30
백준 6593번 - C++  (0) 2022.04.30
백준 5427번 - C++  (0) 2022.04.30