feat: 2025-12-12 baekjoon 10026 문제 풀이

daewook/2025-12-12_baekjoon_10026.cpp

@@ -0,0 +1,103 @@
+#include <iostream>
+#include <queue>
+
+using namespace std;
+
+int n;
+
+int dx[4] = {1, -1, 0, 0};
+int dy[4] = {0, 0, 1, -1};
+
+char ground[100][100];
+bool visited1[100][100];
+bool visited2[100][100];
+
+void bfs(int x, int y, char color) {
+    queue<pair<int, int>> q1;
+    visited1[x][y] = true;
+    q1.push({x, y});
+
+    while (!q1.empty()) {
+        auto [nx, ny] = q1.front();
+        q1.pop();
+
+        for (int i = 0; i < 4; i++) {
+            int cx = nx + dx[i];
+            int cy = ny + dy[i];
+
+            if (cx < 0 || cy < 0 || cx >= n || cy >= n)
+                continue;
+
+            if (!visited1[cx][cy] && ground[cx][cy] == color) {
+                visited1[cx][cy] = true;
+                q1.push({cx, cy});
+            }
+        }
+    }
+}
+
+void bfs2(int x, int y, char color) {
+    queue<pair<int, int>> q2;
+    visited2[x][y] = true;
+    q2.push({x, y});
+
+    while (!q2.empty()) {
+        auto [nx, ny] = q2.front();
+        q2.pop();
+
+        for (int i = 0; i < 4; i++) {
+            int cx = nx + dx[i];
+            int cy = ny + dy[i];
+
+            if (cx < 0 || cy < 0 || cx >= n || cy >= n)
+                continue;
+
+            if (!visited2[cx][cy] && ground[cx][cy] == color) {
+                visited2[cx][cy] = true;
+                q2.push({cx, cy});
+            }
+        }
+    }
+}
+
+int main() {
+    cin >> n;
+
+    for (int i = 0; i < n; i++) {
+        for (int j = 0; j < n; j++) {
+            cin >> ground[i][j];
+        }
+    }
+
+    int n1 = 0;
+    for (int i = 0; i < n; i++) {
+        for (int j = 0; j < n; j++) {
+            if (!visited1[i][j]) {
+                bfs(i, j, ground[i][j]);
+                n1++;
+            }
+        }
+    }
+
+    int n2 = 0;
+
+    // 입력값 수정
+    for (int i = 0; i < n; i++) {
+        for (int j = 0; j < n; j++) {
+            if (ground[i][j] == 'R')
+                ground[i][j] = 'G';
+        }
+    }
+
+    for (int i = 0; i < n; i++) {
+        for (int j = 0; j < n; j++) {
+            if (!visited2[i][j]) {
+                bfs2(i, j, ground[i][j]);
+                n2++;
+            }
+        }
+    }
+
+cout << n1 << " " << n2;
+return 0;
+}