7월 5주차 발제/과제 자료 업로드

_WeeklyChallenges/W32-[Graph(DFSBFS)]/Assignment_BOJ_2468_안전영역.py

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+"""
+BOJ #2468. 안전 영역 (실버2)
+https://www.acmicpc.net/problem/2468
+유형: Graph, DFS, BFS
+"""
+
+"""
+DFS 풀이
+"""
+import sys
+sys.setrecursionlimit(100000)
+
+input = sys.stdin.readline
+
+
+directions = [(-1, 0), (0, 1), (1, 0), (0, -1)]
+def dfs(x, y):
+    visited[x][y] = True
+    for dx, dy in directions:
+        nx, ny = x + dx, y + dy
+        if 0 <= nx < N and 0 <= ny < N and \
+                not visited[nx][ny] and map[nx][ny] > h:
+            dfs(nx, ny)
+
+
+N = int(input())
+map = [list(map(int, input().split())) for _ in range(N)]
+max_height = 0
+
+# 맵 내의 최대값 구하기
+for m in map:
+    max_height = max(max_height, max(m))
+
+answer = 0
+
+for h in range(0, max_height + 1): # 물이 잠기지 않는 상황을 고려하여 0부터 시작한다.
+    visited = [[False for _ in range(N)] for _ in range(N)]
+    count = 0
+
+    for i in range(N):
+        for j in range(N):
+            if not visited[i][j] and map[i][j] > h:
+                dfs(i, j)
+                count += 1
+
+    answer = max(answer, count)
+
+print(answer)
+
+"""
+BFS 풀이
+"""
+from collections import deque
+
+n = int(input())
+graph = []
+maxNum = 0
+
+for i in range(n):
+    graph.append(list(map(int, input().split())))
+    for j in range(n):
+        if graph[i][j] > maxNum:
+            maxNum = graph[i][j]
+
+dx = [0, 0, 1, -1]
+dy = [1, -1, 0, 0]
+
+
+def bfs(a, b, value, visited):
+    q = deque()
+    q.append((a, b))
+    visited[a][b] = 1
+
+    while q:
+        x, y = q.popleft()
+
+        for i in range(4):
+            nx = x + dx[i]
+            ny = y + dy[i]
+            if 0 <= nx < n and 0 <= ny < n:
+                if graph[nx][ny] > value and visited[nx][ny] == 0:
+                    visited[nx][ny] = 1
+                    q.append((nx, ny))
+
+
+result = 0
+for i in range(maxNum):
+    visited = [[0] * n for i in range(n)]
+    cnt = 0
+
+    for j in range(n):
+        for k in range(n):
+            if graph[j][k] > i and visited[j][k] == 0:
+                bfs(j, k, i, visited)
+                cnt += 1
+
+    if result < cnt:
+        result = cnt
+
+print(result)

_WeeklyChallenges/W32-[Graph(DFSBFS)]/README.md

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+## 🚀7월 5주차 (7/29) 스터디 발제 주제: Graph (DFS/BFS)
+> 발제자: 김민정 (@Mingguriguri)
+
+> [!NOTE]
+> 주제: Graph (DFS/BFS)
+
+### 🗂️ 스터디 자료
+- PDF: [바로가기
+](Study_PGS_81302.pdf)
+
+
+### 📖 문제
+- [프로그래머스 #81302. 거리두기 확인하기](https://www.acmicpc.net/problem/1759): Graph (DFS/BFS) / Level 2
+- 정답 코드: [Study_PGS_81302_거리두기확인하기.py](Study_PGS_81302_거리두기확인하기.py)
+
+### 💻 과제
+- [백준 #2468. 안전구역](https://www.acmicpc.net/problem/2468): Graph (DFS/BFS) / 실버1
+- 정답 코드: [Assignment_BOJ_2468_안전구역.py](Assignment_BOJ_2468_안전구역.py)

_WeeklyChallenges/W32-[Graph(DFSBFS)]/Study_PGS_81302_거리두기확인하기.py

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+"""
+PGS #81032. 거리두기 확인하기 (Level 2)
+https://school.programmers.co.kr/learn/courses/30/lessons/81302
+유형: Brute Force, Graph, BFS
+"""
+
+"""
+풀이1: 완전 탐색
+"""
+
+
+def solution(places):
+    def is_safe(place):
+        for i in range(5):
+            for j in range(5):
+                if place[i][j] != 'P':
+                    continue
+
+                for dx, dy in dirs:
+                    ni, nj = i + dx, j + dy
+                    if not (0 <= ni < 5 and 0 <= nj < 5):
+                        continue
+                    if place[ni][nj] != 'P':
+                        continue
+
+                    dist = abs(dx) + abs(dy)
+                    if dist == 1:
+                        return 0  # 거리 1에서 바로 P면 위반
+                    elif dist == 2:
+                        # 파티션 여부 확인
+                        if dx == 0:  # 수평
+                            if place[i][j + dy // 2] != 'X':
+                                return 0
+                        elif dy == 0:  # 수직
+                            if place[i + dx // 2][j] != 'X':
+                                return 0
+                        else:  # 대각선
+                            if place[i][nj] != 'X' or place[ni][j] != 'X':
+                                return 0
+        return 1
+
+    dirs = [
+        (-1, 0), (1, 0), (0, -1), (0, 1),  # 거리 1
+        (-2, 0), (2, 0), (0, -2), (0, 2),  # 일직선 거리 2
+        (-1, -1), (-1, 1), (1, -1), (1, 1)  # 대각선 거리 2
+    ]
+
+    answer = []
+    for place in places:
+        answer.append(is_safe(place))
+
+    return answer
+
+
+"""
+풀이2: BFS
+"""
+from collections import deque
+
+
+def bfs(p):
+    start = []
+
+    for i in range(5):  # 시작점이 되는 P 좌표 구하기
+        for j in range(5):
+            if p[i][j] == 'P':
+                start.append([i, j])
+
+    for s in start:
+        queue = deque([s])  # 큐에 초기값
+        visited = [[0] * 5 for i in range(5)]  # 방문 처리 리스트
+        distance = [[0] * 5 for i in range(5)]  # 경로 길이 리스트
+        visited[s[0]][s[1]] = 1
+
+        while queue:
+            y, x = queue.popleft()
+
+            dx = [-1, 1, 0, 0]  # 좌우
+            dy = [0, 0, -1, 1]  # 상하
+
+            for i in range(4):
+                nx = x + dx[i]
+                ny = y + dy[i]
+
+                if 0 <= nx < 5 and 0 <= ny < 5 and visited[ny][nx] == 0:
+
+                    if p[ny][nx] == 'O':
+                        queue.append([ny, nx])
+                        visited[ny][nx] = 1
+                        distance[ny][nx] = distance[y][x] + 1
+
+                    if p[ny][nx] == 'P' and distance[y][x] <= 1:
+                        return 0
+    return 1
+
+
+def solution(places):
+    answer = []
+
+    for i in places:
+        answer.append(bfs(i))
+
+    return answer