BOJ 10026 - 적록색약

www.acmicpc.net/problem/10026

10026번: 적록색약

 

그냥 BFS 를 두번 쓰면 되는 그런 문제였다.

 

 

- C++

#include <cstring>
#include <iostream>
#include <queue>
 
#define MAX 101
 
using namespace std;
 
int N, answer1, answer2;
int dy[4] = {-1, 1, 0, 0};
int dx[4] = {0, 0, -1, 1};
char adj[MAX][MAX];
bool visited[MAX][MAX];
 
void bfs1(int r, int c) {
    queue<pair<int, int>> q;
    q.push({r, c});
    visited[r][c] = true;
    
    while(!q.empty()) {
        int y = q.front().first, x = q.front().second;
        char ch = adj[y][x];
        q.pop();
        
        for (int i = 0; i < 4; ++i) {
            int ny = y + dy[i], nx = x + dx[i];
            
            if (visited[ny][nx] || ny < 0 || ny >= N || nx < 0 || nx >= N) continue;
            
            if (adj[ny][nx] == ch) {
                visited[ny][nx] = true;
                q.push({ny, nx});
            }
        }
    }
}
 
void bfs2(int r, int c) {
    queue<pair<int, int>> q;
    q.push({r, c});
    visited[r][c] = true;
    
    while(!q.empty()) {
        int y = q.front().first, x = q.front().second;
        char ch = adj[y][x];
        q.pop();
        
        for (int i = 0; i < 4; ++i) {
            int ny = y + dy[i], nx = x + dx[i];
            
            if (visited[ny][nx] || ny < 0 || ny >= N || nx < 0 || nx >= N) continue;
            
            if (ch == 'R' || ch == 'G') {
                if (adj[ny][nx] == 'R' || adj[ny][nx] == 'G') {
                    visited[ny][nx] = true;
                    q.push({ny, nx});
                }
            } else {
                if (adj[ny][nx] == ch) {
                    visited[ny][nx] = true;
                    q.push({ny, nx});
                }
            }
        }
    }
}
 
int main() {
    cin.tie(0); ios_base::sync_with_stdio(0);
    
    cin >> N;
    
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < N; ++j)
            cin >> adj[i][j];
    
    
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            if (!visited[i][j]) {
                bfs1(i, j);
                answer1++;
            }
        }
    }
    
    memset(visited, false, sizeof(visited));
    
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            if (!visited[i][j]) {
                bfs2(i, j);
                answer2++;
            }
        }
    }
    
    cout << answer1 << " " << answer2;
    
    return 0;
}