[class3] (백준 10026) 적록색약

문제

적록색약은 빨간색과 초록색의 차이를 거의 느끼지 못한다. 따라서, 적록색약인 사람이 보는 그림은 아닌 사람이 보는 그림과는 좀 다를 수 있다.

크기가 N×N인 그리드의 각 칸에 R(빨강), G(초록), B(파랑) 중 하나를 색칠한 그림이 있다. 그림은 몇 개의 구역으로 나뉘어져 있는데, 구역은 같은 색으로 이루어져 있다. 또, 같은 색상이 상하좌우로 인접해 있는 경우에 두 글자는 같은 구역에 속한다. (색상의 차이를 거의 느끼지 못하는 경우도 같은 색상이라 한다)

예를 들어, 그림이 아래와 같은 경우에

RRRBB
GGBBB
BBBRR
BBRRR
RRRRR

적록색약이 아닌 사람이 봤을 때 구역의 수는 총 4개이다. (빨강 2, 파랑 1, 초록 1) 하지만, 적록색약인 사람은 구역을 3개 볼 수 있다. (빨강-초록 2, 파랑 1)

그림이 입력으로 주어졌을 때, 적록색약인 사람이 봤을 때와 아닌 사람이 봤을 때 구역의 수를 구하는 프로그램을 작성하시오.

입력

첫째 줄에 N이 주어진다. (1 ≤ N ≤ 100)

둘째 줄부터 N개 줄에는 그림이 주어진다.

출력

적록색약이 아닌 사람이 봤을 때의 구역의 개수와 적록색약인 사람이 봤을 때의 구역의 수를 공백으로 구분해 출력한다.

제한

 

예제 입력 1

5
RRRBB
GGBBB
BBBRR
BBRRR
RRRRR

예제 출력 1

4 3

힌트

 

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

출처

Olympiad > USA Computing Olympiad > 2013-2014 Season > USACO March 2014 Contest > Bronze 3번

• 문제를 번역한 사람: baekjoon
• 어색한 표현을 찾은 사람: corea

#include <iostream> 
#include <cstring>

using namespace std; 
int n, cnt;
char arr[100][100];
bool visit[100][100] = { false };

void DFS(int x, int y) 
{
	int dx[4] = { 0,0,-1,1 };
	int dy[4] = { -1,1,0,0 };

	visit[x][y] = true;
	for (int i = 0; i < 4; i++) 
	{
		int tempX = x + dx[i];
		int tempY = y + dy[i];

		if (tempX < 0 || tempX >= n || tempY < 0 || tempY >= n) 
			continue;
		if (!visit[tempX][tempY] && arr[x][y] == arr[tempX][tempY]) 
		{
			DFS(tempX, tempY);
		}
	}
}
int main() 
{ 
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin >> n;
	memset(visit, false, sizeof(visit));
	for (int i = 0; i < n; i++) 
	{
		cin >> arr[i];
	}

	for (int i = 0; i < n; i++) 
	{
		for (int j = 0; j < n; j++) 
		{
			if (!visit[i][j]) 
			{
				DFS(i, j);
				cnt++;
			}
		}
	}
	cout << cnt << ' ';

	for (int i = 0; i < n; i++) 
	{
		for (int j = 0; j < n; j++)
		{
			if (arr[i][j] == 'G')
			{
				arr[i][j] = 'R';
			}
			else
			{
				arr[i][j] = arr[i][j];
			}
		}
	}
	memset(visit, false, sizeof(visit));

	cnt = 0;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (!visit[i][j])
			{
				DFS(i, j);
				cnt++;
			}
		}
	}
	cout << cnt;

	return 0;
}