C++(씨쁠쁠)(cplusplus)-백준(baekjoon)(BaekJoon)코딩 10026번:적록색약 답

R,G,B각각에 대해 DFS를 시행한다. R:1, G:2, B:3

#include <iostream>
using namespace std;
char k;
int h, w, a[101][101], b[101][101], acheck[101][101], bcheck[101][101], n, acnt,bcnt, dh[] = { 0,0,-1,1 }, dw[] = { -1,1,0,0 };//a는 정상인 b는 색약
void aDFS(int h, int w, int color,int acnt)
{
    acheck[h][w] = color;
    for (int i = 0; i < 4; i++)
    {
        int nh = h + dh[i];
        int nw = w + dw[i];
        if (0 <= nh && nh < n && 0 <= nw && nw <= n)
        {
            if (a[nh][nw] == color && acheck[nh][nw] == 0)
                aDFS(nh, nw, color,acnt);
            
        }
    }
}
void bDFS(int h, int w, int color, int bcnt)
{
    bcheck[h][w] = color;
    for (int i = 0; i < 4; i++)
    {
        int nh = h + dh[i];
        int nw = w + dw[i];
        if (0 <= nh && nh < n && 0 <= nw && nw <= n)
        {
            if (b[nh][nw] == color && bcheck[nh][nw] == 0)
                bDFS(nh, nw, color, bcnt);
            
        
        }
    }
}
int main() {
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            cin >> k;
            
            switch (k)
            {
            case 'R': a[i][j] = 1; b[i][j] = 1; break;
            case 'G': a[i][j] = 2; b[i][j] = 1; break;
            case 'B':a[i][j] = 3; b[i][j] = 2; break;
            }
        }
        cin.ignore();
    }
    /*for (int i = 0; i<n; i++,puts(""))
        for (int j = 0; j < n; j++)
        {
            cout << a[i][j] << ' ';
        }
    cout << '\n';
    for (int i = 0; i<n; i++, puts(""))
        for (int j = 0; j < n; j++)
        {
            cout << b[i][j] << ' ';
        }*/
    for(int i = 0; i<n; i++)
        for (int j = 0; j < n; j++)
        {
            if (a[i][j] == 1 && acheck[i][j] == 0 &&(acheck[i][j] !=2 || acheck[i][j] != 3)) { aDFS(i, j, 1, ++acnt); }
            if (a[i][j] == 2 && acheck[i][j] == 0 && (acheck[i][j] != 1 || acheck[i][j] != 3)) { aDFS(i, j, 2, ++acnt); }
            if (a[i][j] == 3 && acheck[i][j] == 0 && (acheck[i][j] != 1 || acheck[i][j] != 2)) { aDFS(i, j, 3, ++acnt); }
            if (b[i][j] == 1 && bcheck[i][j] == 0 && (bcheck[i][j] != 2 || bcheck[i][j] != 3)) { bDFS(i, j, 1, ++bcnt); }
            if (b[i][j] == 2 && bcheck[i][j] == 0 && (bcheck[i][j] != 1 || bcheck[i][j] != 3)) { bDFS(i, j, 2, ++bcnt); }
        }
    cout << acnt << ' ' << bcnt;
}