17-H0ngJu #195

H0ngJu · 2024-05-22T09:01:45Z

🔗 문제 링크

적록색약

✔️ 소요된 시간

1H 20M

✨ 수도 코드

문제 요약
정상인과, 적녹색약인 사람이 주어진 그림을 보았을 때, 각각 몇 개의 구역으로 보이는지 반환하라@!

bfs+백트래킹이라고 생각을 했고

주어진 그림에서, R 구역의 수, G 구역의 수, B 구역의 수를 각각 따로 구한 뒤에

정상인이 바라보는 구역의 개수는 R + G + B,

적녹색약인 경우 R + B로 단순하게 구하려고 했습니다

적색과 녹색을 구분하지 않아도 되므로 G를 안 세면 된다고 생각했습니다.

즉, 주어진 그림에서 R을 G로 바꾸지 않아도 가능할 것이라고 예상했습니다.

코드보기

import sys

def input(): return sys.stdin.readline().rstrip()

N = int(input())
arr = [[i for i in input()]for _ in range(N)]
visited = [[0] * N for _ in range(N)]
direc = [(1,0), (-1,0), (0,1), (0,-1)]
result = [0, 0, 0] # R G B 구역 수

def dfs(x, y, color):
    visited[x][y] = 1
    for dx, dy in direc:
        nx = dx + x
        ny = dy + y
        if 0 <= nx < N and 0 <= ny < N and not visited[nx][ny] and arr[nx][ny] == color:
            dfs(nx, ny, color)

def solution(color):
    cnt = 0
    for i in range(N):
        for j in range(N):
            if not visited[i][j] and arr[i][j] == color:
                dfs(i,j,color)
                cnt += 1

    return cnt

result[0] = solution("R")
visited = [[0] * N for _ in range(N)]
result[1] = solution("G")
visited = [[0] * N for _ in range(N)]
result[2] = solution("B")

print(result[0]+result[1]+result[2], result[0]+result[2], end=" ")

오 쉬운데? 하고 제출했더니 '틀렸습니다'가 떴습니다 ㅎ;;

아래의 경우가 반례입니다.

정상인의 경우 R = 2, G = 2, B = 0이므로, 4입니다.

단순히 적녹색약인 경우 R + B로 계산하게 되면, 2가 나옵니다.

하지만, 실제로 적녹색약이 바라보는 구역의 개수는 1입니다.

R이랑 G 바꾸지 않아도 적녹색약의 구역 수를 찾을 수 있을거라는 생각과는 달리,

위의 경우에서 결국 R과 G를 바꿔준 후, 구역 수를 세야한다는 것을 깨닫고 로직을 고쳤습니다 !

최종 수도 코드

정상인의 경우
1. 아직 방문하지 않은 노드라면 q에 넣고 visited 갱신
2. q가 빌 동안 아래를 수행
  1. 현재 위치 pop
  2. 만약 다음 위치(상하좌우)가 색이 같으면 q에 넣고 다시 visited 갱신
적녹 색약의 경우
0. 먼저 R을 G로 바꿔주고 정상인과 같은 로직 수행

import sys
sys.setrecursionlimit(10**6)
from collections import deque

def input(): return sys.stdin.readline().rstrip()

N = int(input())
arr = [[i for i in input()]for _ in range(N)]
visited = [[0] * N for _ in range(N)]
direc = [(1,0), (-1,0), (0,1), (0,-1)]
result = [0, 0]

def bfs(x, y, color):
    q = deque([(x,y)])
    visited[x][y] = 1
    while q:
        cx, cy = q.popleft()
        for dx, dy in direc:
            nx = dx + cx
            ny = dy + cy
            if 0 <= nx < N and 0 <= ny < N and not visited[nx][ny] and arr[nx][ny] == color:
                q.append((nx, ny))
                visited[nx][ny] = 1


for i in range(N):
    for j in range(N):
        if not visited[i][j]:
            bfs(i,j,arr[i][j])
            result[0] += 1

visited = [[0] * N for _ in range(N)]

for i in range(N):
    for j in range(N):
        if arr[i][j] == "R":
            arr[i][j] = "G"

for i in range(N):
    for j in range(N):
        if not visited[i][j]:
            bfs(i,j,arr[i][j])
            result[1] += 1


print(result[0], result[1], end=" ")

📚 새롭게 알게된 내용

SeongHoonC

아.. G 를 R 로 바꿀 생각을 못하고 bfs 코드를 두개 짜버렸네요..ㅋㅋ
홍주님 방법으로 다시 작성해봤습니다..!

코드 길이도 시간도 더 줄었네요!

import java.io.BufferedReader
import java.io.InputStreamReader

private lateinit var visited: Array<Array<Boolean>>
private lateinit var graph: Array<Array<String>>

private val dxdy = listOf((0 to 1), (0 to -1), (1 to 0), (-1 to 0))
fun main() {
    val br = BufferedReader(InputStreamReader(System.`in`))
    val n = br.readLine().toInt()

    graph = Array(n) { Array(n) { "" } }

    repeat(n) {
        val line = br.readLine().chunked(1)
        line.forEachIndexed { index, value -> graph[it][index] = value }
    }

    visited = Array(n) { Array(n) { false } }
    var count = 0
    var countRG = 0
    for (x in 0 until n) {
        for (y in 0 until n) {
            if (bfs(x, y, n)) count++
        }
    }
    visited = Array(n) { Array(n) { false } }
    for (x in 0 until n) {
        for (y in 0 until n) {
            if (bfsRG(x, y, n)) countRG++
        }
    }

    println("$count $countRG")
}

private fun bfs(x: Int, y: Int, n: Int): Boolean {
    if (visited[x][y]) {
        return false
    }
    val q = ArrayDeque<Pair<Int, Int>>()
    q.add(x to y)
    val color = graph[x][y]
    while (q.isNotEmpty()) {
        val now = q.removeFirst()
        for (i in 0..3) {
            val nextX = now.first + dxdy[i].first
            val nextY = now.second + dxdy[i].second
            if (nextX >= n || nextY >= n || nextX < 0 || nextY < 0) {
                continue
            }
            val nextColor = graph[nextX][nextY]
            if (visited[nextX][nextY]) {
                continue
            }
            if (color != nextColor) {
                continue
            }
            visited[nextX][nextY] = true
            q.add(nextX to nextY)
        }
    }
    return true
}

private fun bfsRG(x: Int, y: Int, n: Int): Boolean {
    if (visited[x][y]) {
        return false
    }
    val q = ArrayDeque<Pair<Int, Int>>()
    q.add(x to y)
    val color = graph[x][y]
    while (q.isNotEmpty()) {
        val now = q.removeFirst()
        for (i in 0..3) {
            val nextX = now.first + dxdy[i].first
            val nextY = now.second + dxdy[i].second
            if (nextX >= n || nextY >= n || nextX < 0 || nextY < 0) {
                continue
            }
            val nextColor = graph[nextX][nextY]
            if (visited[nextX][nextY]) {
                continue
            }
            if (color != nextColor && (color == "B" || nextColor == "B")) {
                continue
            }
            visited[nextX][nextY] = true
            q.add(nextX to nextY)
        }
    }
    return true
}

G 를 R 로 바꾸고 실행 : bfs 코드는 한개

import java.io.BufferedReader
import java.io.InputStreamReader

private lateinit var visited: Array<Array<Boolean>>
private lateinit var graph: Array<Array<String>>

private val dxdy = listOf((0 to 1), (0 to -1), (1 to 0), (-1 to 0))
fun main() {
    val br = BufferedReader(InputStreamReader(System.`in`))
    val n = br.readLine().toInt()

    graph = Array(n) { Array(n) { "" } }

    repeat(n) {
        val line = br.readLine().chunked(1)
        line.forEachIndexed { index, value -> graph[it][index] = value }
    }

    visited = Array(n) { Array(n) { false } }
    var count = 0

    for (x in 0 until n) {
        for (y in 0 until n) {
            if (bfs(x, y, n)) count++
        }
    }

    for (x in 0 until n) {
        for (y in 0 until n) {
            if (graph[x][y] == "G") {
                graph[x][y] = "R"
            }
        }
    }

    var countRG = 0
    visited = Array(n) { Array(n) { false } }
    for (x in 0 until n) {
        for (y in 0 until n) {
            if (bfs(x, y, n)) countRG++
        }
    }

    println("$count $countRG")
}

private fun bfs(x: Int, y: Int, n: Int): Boolean {
    if (visited[x][y]) {
        return false
    }
    val q = ArrayDeque<Pair<Int, Int>>()
    q.add(x to y)
    val color = graph[x][y]
    while (q.isNotEmpty()) {
        val now = q.removeFirst()
        for (i in 0..3) {
            val nextX = now.first + dxdy[i].first
            val nextY = now.second + dxdy[i].second
            if (nextX >= n || nextY >= n || nextX < 0 || nextY < 0) {
                continue
            }
            val nextColor = graph[nextX][nextY]
            if (visited[nextX][nextY]) {
                continue
            }
            if (color != nextColor) {
                continue
            }
            visited[nextX][nextY] = true
            q.add(nextX to nextY)
        }
    }
    return true
}

9kyo-hwang

제 경우엔 1. 일반인용 보드 2. 적록색약용 보드 2가지를 만들어서 각 보드에 대해 dfs를 수행하는 식으로 풀었었네요 :)
적록색약일 경우 R이나 G를 다른 한쪽으로 통일시키는 게 이 문제를 푸는 key라고 생각됩니다 :)

#include <iostream>
#include <vector>

using namespace std;

int N;

void dfs(vector<vector<char>> &grid, int x, int y, char color) {
  if (x < 0 || x >= N || y < 0 || y >= N) {
    return;
  }

  if (grid[x][y] != color) {
    return;
  }

  grid[x][y] = 'A';

  dfs(grid, x - 1, y, color);
  dfs(grid, x, y + 1, color);
  dfs(grid, x + 1, y, color);
  dfs(grid, x, y - 1, color);
 }

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);

  cin >> N;

  vector<vector<char>> grid1, grid2;
  grid1.resize(N, vector<char>(N));
  grid2.resize(N, vector<char>(N));

  char ch;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      cin >> ch;
      grid1[i][j] = ch;
      
      if (ch == 'R') {
        ch = 'G';
      }
      grid2[i][j] = ch;
    }
  }

  int count1 = 0, count2 = 0;

  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (grid1[i][j] != 'A') {
        dfs(grid1, i, j, grid1[i][j]);
        count1 += 1;
      }

      if (grid2[i][j] != 'A') {
        dfs(grid2, i, j, grid2[i][j]);
        count2 += 1;
      }
    }
  }

  cout << count1 << " " << count2;

  return 0;
}

tgyuuAn

BFS/DFS로 풀릴 거 알고 있었는데 유니온 파인드로도 풀릴 것 같아서 트라이 해봤는데 왜 안되는 지 모르겠네요 .......... 도와주세요 ㅠㅠㅠㅠㅠㅠㅠ

from collections import Counter
import sys

def input(): return sys.stdin.readline().rstrip()

N = int(input())

board = []

def find_parent(graph, element):
    if graph[element] == element: return element
    parent = graph[element]
    graph[element] = find_parent(graph, parent)
    return graph[element]

def union(graph, first, second):
    x = find_parent(graph, first)
    y = find_parent(graph, second)
    if x == y: return

    x, y = min(x, y), max(x, y)
    graph[y] = x
    return

normal_parent = []
color_blindness_parent = []

idx = 0
for _ in range(N):
    row = list(input())
    
    for element in row:
        normal_parent.append(idx)
        color_blindness_parent.append(idx)
        idx += 1

    board.append(row)

dx = [0, 0, -1, 1]
dy = [-1, 1, 0, 0]

for row in range(N):
    for col in range(N):

        for dir in range(4):
            new_row = row + dy[dir]
            new_col = col + dx[dir]

            if new_row < 0 or new_row >= N: continue
            if new_col < 0 or new_col >= N: continue

            if board[row][col] == board[new_row][new_col]:
                union(normal_parent, ((N * row) + col), ((N * new_row) + new_col))

            if board[row][col] in ("R", "G") and board[new_row][new_col] in ("R", "G") or board[row][col] == board[new_row][new_col] == "B":
                union(color_blindness_parent, (N * row + col), (N * new_row + new_col))

print(len(Counter(normal_parent).keys()), len(Counter(color_blindness_parent).keys()))

결국 유니온 파인드 포기하고 BFS로 틀었씁니다..

from collections import deque
import sys

def input(): return sys.stdin.readline().rstrip()

N = int(input())

idx = 0
board = []
for _ in range(N):
    board.append(list(input()))

dx = [0, 0, -1, 1]
dy = [-1, 1, 0, 0]

visited = set()
deq = deque()
normal_answer = 0
for row in range(N):
    for col in range(N):
        if (row, col) in visited: continue
    
        normal_answer += 1
        visited.add((row, col))
        deq.append((row, col))

        while deq:
            now_row, now_col = deq.popleft()

            for dir in range(4):
                new_row = now_row + dy[dir]
                new_col = now_col + dx[dir]

                if new_row < 0 or new_row >= N: continue
                if new_col < 0 or new_col >= N: continue
                if (new_row, new_col) in visited: continue

                if board[row][col] == board[new_row][new_col]:
                    deq.append((new_row, new_col))
                    visited.add((new_row, new_col))

visited = set()
deq = deque()
color_blindness_answer = 0
for row in range(N):
    for col in range(N):
        if (row, col) in visited: continue

        color_blindness_answer += 1
        visited.add((row, col))
        deq.append((row, col))

        while deq:
            now_row, now_col = deq.popleft()

            for dir in range(4):
                new_row = now_row + dy[dir]
                new_col = now_col + dx[dir]

                if new_row < 0 or new_row >= N: continue
                if new_col < 0 or new_col >= N: continue
                if (new_row, new_col) in visited: continue

                if board[row][col] in ("R", "G") and board[new_row][new_col] in ("R", "G") or board[row][col] == board[new_row][new_col] == "B":
                    deq.append((new_row, new_col))
                    visited.add((new_row, new_col))

print(normal_answer, color_blindness_answer)

와 근데 R이랑 G 똑같이 맞추고 다시 돌리는 거 뤠전두다 뤠전두

tgyuuAn · 2024-05-24T02:39:10Z

H0ngJu/BFS/적록색약.py

+for i in range(N):
+    for j in range(N):
+        if arr[i][j] == "R":
+            arr[i][j] = "G"


와 R이랑 G 똑같이 만드는 거 미쳤다..

alstjr7437 · 2024-05-25T12:55:53Z

코드가 조금 더럽게 나왔네요 저는...
근데 전체 로직이 똑같아서 신기했습니다!!!

from collections import deque
import sys

dx = [0,0,-1,1]
dy = [1,-1,0,0]

input = sys.stdin.readline

n = int(input())
visited = [[False] * n for _ in range(n)] 
graph = [list(input().rstrip()) for _ in range(n)]
queue = deque()
result = 0


def bfs():
    global result 
    while queue:
        x, y, color = queue.popleft()
        for i in range(4):
            nx, ny= x + dx[i], y + dy[i]
            if nx < 0 or nx >= n or ny < 0 or ny >= n:
                continue
            if color == graph[ny][nx] and not visited[ny][nx] :
                queue.append((nx, ny, color))
                visited[ny][nx] = True
    result += 1

def resultCount():
    global result 
    for y in range(n):
        for x in range(n):
            if not visited[y][x] :
                queue.append((x, y, graph[y][x]))
                visited[y][x] = True
                bfs()
    print(result, end=" ")
    result = 0

# 적록색약 아닐때 출력
resultCount()

# 적록색약 적용 부분
for y in range(n):
    for x in range(n):
        visited[y][x] = False
        if graph[y][x] == "G":
            graph[y][x] = "R"

# 적록색약 일때 출력
resultCount()

alstjr7437 · 2024-05-25T12:50:27Z

H0ngJu/BFS/적록색약.py

+            bfs(i,j,arr[i][j])
+            result[0] += 1
+
+visited = [[0] * N for _ in range(N)]


R,G을 다시 바꾸는게 같아서 확인해보니 어짜피 모든 graph를 도는데
그래서 저는 이부분에서 그냥 visited를 새로 만들지 않고 False로 초기화를 해줬습니다!

for y in range(n): for x in range(n): visited[y][x] = False if graph[y][x] == "G": graph[y][x] = "R"

alstjr7437 · 2024-05-25T12:51:16Z

H0ngJu/BFS/적록색약.py

+    for j in range(N):
+        if not visited[i][j]:
+            bfs(i,j,arr[i][j])
+            result[0] += 1


result 배열 두개 만들어두고 각 부분에 맞게 추가하는 부분이 있었군요 ㄷㄷㄷㄷㄷㄷㄷ
저는 출력하고 다시 초기화 해주는 형식으로 했는데 이 방법이 더 좋은 것 같네요!

H0ngJu · 2024-05-26T10:08:00Z

BFS/DFS로 풀릴 거 알고 있었는데 유니온 파인드로도 풀릴 것 같아서 트라이 해봤는데 왜 안되는 지 모르겠네요 .......... 도와주세요 ㅠㅠㅠㅠㅠㅠㅠ

from collections import Counter
import sys

def input(): return sys.stdin.readline().rstrip()

N = int(input())

board = []

def find_parent(graph, element):
    if graph[element] == element: return element
    parent = graph[element]
    graph[element] = find_parent(graph, parent)
    return graph[element]

def union(graph, first, second):
    x = find_parent(graph, first)
    y = find_parent(graph, second)
    if x == y: return

    x, y = min(x, y), max(x, y)
    graph[y] = x
    return

normal_parent = []
color_blindness_parent = []

idx = 0
for _ in range(N):
    row = list(input())
    
    for element in row:
        normal_parent.append(idx)
        color_blindness_parent.append(idx)
        idx += 1

    board.append(row)

dx = [0, 0, -1, 1]
dy = [-1, 1, 0, 0]

for row in range(N):
    for col in range(N):

        for dir in range(4):
            new_row = row + dy[dir]
            new_col = col + dx[dir]

            if new_row < 0 or new_row >= N: continue
            if new_col < 0 or new_col >= N: continue

            if board[row][col] == board[new_row][new_col]:
                union(normal_parent, ((N * row) + col), ((N * new_row) + new_col))

            if board[row][col] in ("R", "G") and board[new_row][new_col] in ("R", "G") or board[row][col] == board[new_row][new_col] == "B":
                union(color_blindness_parent, (N * row + col), (N * new_row + new_col))

print(len(Counter(normal_parent).keys()), len(Counter(color_blindness_parent).keys()))

결국 유니온 파인드 포기하고 BFS로 틀었씁니다..

from collections import deque
import sys

def input(): return sys.stdin.readline().rstrip()

N = int(input())

idx = 0
board = []
for _ in range(N):
    board.append(list(input()))

dx = [0, 0, -1, 1]
dy = [-1, 1, 0, 0]

visited = set()
deq = deque()
normal_answer = 0
for row in range(N):
    for col in range(N):
        if (row, col) in visited: continue
    
        normal_answer += 1
        visited.add((row, col))
        deq.append((row, col))

        while deq:
            now_row, now_col = deq.popleft()

            for dir in range(4):
                new_row = now_row + dy[dir]
                new_col = now_col + dx[dir]

                if new_row < 0 or new_row >= N: continue
                if new_col < 0 or new_col >= N: continue
                if (new_row, new_col) in visited: continue

                if board[row][col] == board[new_row][new_col]:
                    deq.append((new_row, new_col))
                    visited.add((new_row, new_col))

visited = set()
deq = deque()
color_blindness_answer = 0
for row in range(N):
    for col in range(N):
        if (row, col) in visited: continue

        color_blindness_answer += 1
        visited.add((row, col))
        deq.append((row, col))

        while deq:
            now_row, now_col = deq.popleft()

            for dir in range(4):
                new_row = now_row + dy[dir]
                new_col = now_col + dx[dir]

                if new_row < 0 or new_row >= N: continue
                if new_col < 0 or new_col >= N: continue
                if (new_row, new_col) in visited: continue

                if board[row][col] in ("R", "G") and board[new_row][new_col] in ("R", "G") or board[row][col] == board[new_row][new_col] == "B":
                    deq.append((new_row, new_col))
                    visited.add((new_row, new_col))

print(normal_answer, color_blindness_answer)

와 근데 R이랑 G 똑같이 맞추고 다시 돌리는 거 뤠전두다 뤠전두

유니온파인드 코드 반례 찾았습니다

20
BBBBBRRRRRRRRRRRBBBB
BBBBBRRRRRRRRRRRBBBB
RBBBBBRRRRRRRRRRBBBB
RRRBBBBRRRRRRRRRBBBB
RRRBBBBRRRRRRRRRRBRB
GRRBBBBRRRRRRRRRRBRR
GGRRRRBBBRRRRRRRRBBB
GGGRRRBBBRRRRRRRRBBB
RRGGGGBBBRRRRRRRRBBB
BBGGGGBBBBRRRRRRRBBB
BBGGGGGBBBRRRRRRRBBB
GBGGGGGBRRRRRRRRRBBB
GGGGGGGGRRRRRRRRRBBB
GGGGGGGGGRRRRRRRRBBB
GGGGGGGGGGRRRRRRRBBB
RRGGGGGGGGGGRRRRRRBB
RRGGGGGGGGGGGGGGGRBB
RRRGGGGGGGGGGGGGGRBB
GGGGGGGBGGGGGGGGGGBB
RRRRGGGGGGGGGGGGGGGG

를 입력하면 답은 11 6인데 12 7이 나오네용

하 근데 왜 안되는지는 저도 아직 못찾겠네요.. 쫌 더 생각해보겠슴둥

=======
@tgyuuAn 찾았습니다 ㅎㅎ

for i in range(len(normal_parent)):
    print(i, " : ", normal_parent[i])

를 돌리면 아래처럼 나와요

그런데 실제로 361번째 값을 보면 부모가 100이어야 하는데 부모가 360으로 되어 있어요(0~400번기준으로 index를 잡은 경우입니다)

그래서

print 출력해서 돌려보니까
graph[360] 값이 100으로 변경되기 전에 이미 graph[361]을 360으로 바꿨기 때문에 graph[361] 100으로 업데이트가 안된 것 같아요!!

그래서

최종 print 문을 출력하기 전에, 한번더 find_parent()를 돌려서 부모노드를 가르키도록 다시 업데이트 해주어야 할 것 같습니다....

애렵네요
유니온 파인드가 원래 이런 경우를 다 생각해주어야 할까요................???

tgyuuAn · 2024-05-27T02:45:29Z

BFS/DFS로 풀릴 거 알고 있었는데 유니온 파인드로도 풀릴 것 같아서 트라이 해봤는데 왜 안되는 지 모르겠네요 .......... 도와주세요 ㅠㅠㅠㅠㅠㅠㅠ
from collections import Counter

import sys

def input(): return sys.stdin.readline().rstrip()

N = int(input())

board = []

def find_parent(graph, element):
if graph[element] == element: return element
parent = graph[element]
graph[element] = find_parent(graph, parent)
return graph[element]
def union(graph, first, second):
x = find_parent(graph, first)
y = find_parent(graph, second)
if x == y: return
x, y = min(x, y), max(x, y)
graph[y] = x
return
normal_parent = []

color_blindness_parent = []

idx = 0

for _ in range(N):
row = list(input())
for element in row:
    normal_parent.append(idx)
    color_blindness_parent.append(idx)
    idx += 1
board.append(row)
dx = [0, 0, -1, 1]

dy = [-1, 1, 0, 0]

for row in range(N):
for col in range(N):
    for dir in range(4):
        new_row = row + dy[dir]
        new_col = col + dx[dir]
        if new_row < 0 or new_row >= N: continue
        if new_col < 0 or new_col >= N: continue
        if board[row][col] == board[new_row][new_col]:
            union(normal_parent, ((N * row) + col), ((N * new_row) + new_col))
        if board[row][col] in ("R", "G") and board[new_row][new_col] in ("R", "G") or board[row][col] == board[new_row][new_col] == "B":
            union(color_blindness_parent, (N * row + col), (N * new_row + new_col))
print(len(Counter(normal_parent).keys()), len(Counter(color_blindness_parent).keys()))
결국 유니온 파인드 포기하고 BFS로 틀었씁니다..
from collections import deque

import sys

def input(): return sys.stdin.readline().rstrip()

N = int(input())

idx = 0

board = []

for _ in range(N):
board.append(list(input()))
dx = [0, 0, -1, 1]

dy = [-1, 1, 0, 0]

visited = set()

deq = deque()

normal_answer = 0

for row in range(N):
for col in range(N):
    if (row, col) in visited: continue
    normal_answer += 1
    visited.add((row, col))
    deq.append((row, col))
    while deq:
        now_row, now_col = deq.popleft()
        for dir in range(4):
            new_row = now_row + dy[dir]
            new_col = now_col + dx[dir]
            if new_row < 0 or new_row >= N: continue
            if new_col < 0 or new_col >= N: continue
            if (new_row, new_col) in visited: continue
            if board[row][col] == board[new_row][new_col]:
                deq.append((new_row, new_col))
                visited.add((new_row, new_col))
visited = set()

deq = deque()

color_blindness_answer = 0

for row in range(N):
for col in range(N):
    if (row, col) in visited: continue
    color_blindness_answer += 1
    visited.add((row, col))
    deq.append((row, col))
    while deq:
        now_row, now_col = deq.popleft()
        for dir in range(4):
            new_row = now_row + dy[dir]
            new_col = now_col + dx[dir]
            if new_row < 0 or new_row >= N: continue
            if new_col < 0 or new_col >= N: continue
            if (new_row, new_col) in visited: continue
            if board[row][col] in ("R", "G") and board[new_row][new_col] in ("R", "G") or board[row][col] == board[new_row][new_col] == "B":
                deq.append((new_row, new_col))
                visited.add((new_row, new_col))
print(normal_answer, color_blindness_answer)
와 근데 R이랑 G 똑같이 맞추고 다시 돌리는 거 뤠전두다 뤠전두

유니온파인드 코드 반례 찾았습니다
20

BBBBBRRRRRRRRRRRBBBB

BBBBBRRRRRRRRRRRBBBB

RBBBBBRRRRRRRRRRBBBB

RRRBBBBRRRRRRRRRBBBB

RRRBBBBRRRRRRRRRRBRB

GRRBBBBRRRRRRRRRRBRR

GGRRRRBBBRRRRRRRRBBB

GGGRRRBBBRRRRRRRRBBB

RRGGGGBBBRRRRRRRRBBB

BBGGGGBBBBRRRRRRRBBB

BBGGGGGBBBRRRRRRRBBB

GBGGGGGBRRRRRRRRRBBB

GGGGGGGGRRRRRRRRRBBB

GGGGGGGGGRRRRRRRRBBB

GGGGGGGGGGRRRRRRRBBB

RRGGGGGGGGGGRRRRRRBB

RRGGGGGGGGGGGGGGGRBB

RRRGGGGGGGGGGGGGGRBB

GGGGGGGBGGGGGGGGGGBB

RRRRGGGGGGGGGGGGGGGG
를 입력하면 답은 11 6인데 12 7이 나오네용

하 근데 왜 안되는지는 저도 아직 못찾겠네요.. 쫌 더 생각해보겠슴둥

=======

@tgyuuAn 찾았습니다 ㅎㅎ
for i in range(len(normal_parent)):

    print(i, " : ", normal_parent[i])
를 돌리면 아래처럼 나와요

그런데 실제로 361번째 값을 보면 부모가 100이어야 하는데 부모가 360으로 되어 있어요(0~400번기준으로 index를 잡은 경우입니다)

그래서

print 출력해서 돌려보니까

graph[360] 값이 100으로 변경되기 전에 이미 graph[361]을 360으로 바꿨기 때문에 graph[361] 100으로 업데이트가 안된 것 같아요!!

그래서

최종 print 문을 출력하기 전에, 한번더 find_parent()를 돌려서 부모노드를 가르키도록 다시 업데이트 해주어야 할 것 같습니다....

애렵네요

유니온 파인드가 원래 이런 경우를 다 생각해주어야 할까요................???

대박적이네뇨.. 와퐁쥬 대박...

생각해보니 마지막에 find_parent() 돌려주어서 변경된 진짜 부모를 찾아줘야해요

저렇게 하묜 시간안에 세잎 되나요?!?!?

H0ngJu · 2024-05-27T04:05:20Z

BFS/DFS로 풀릴 거 알고 있었는데 유니온 파인드로도 풀릴 것 같아서 트라이 해봤는데 왜 안되는 지 모르겠네요 .......... 도와주세요 ㅠㅠㅠㅠㅠㅠㅠ

from collections import Counter

import sys

def input(): return sys.stdin.readline().rstrip()

N = int(input())

board = []

def find_parent(graph, element):
if graph[element] == element: return element
parent = graph[element]
graph[element] = find_parent(graph, parent)
return graph[element]
def union(graph, first, second):
x = find_parent(graph, first)
y = find_parent(graph, second)
if x == y: return
x, y = min(x, y), max(x, y)
graph[y] = x
return
normal_parent = []

color_blindness_parent = []

idx = 0

for _ in range(N):
row = list(input())
for element in row:
    normal_parent.append(idx)
    color_blindness_parent.append(idx)
    idx += 1
board.append(row)
dx = [0, 0, -1, 1]

dy = [-1, 1, 0, 0]

for row in range(N):
for col in range(N):
    for dir in range(4):
        new_row = row + dy[dir]
        new_col = col + dx[dir]
        if new_row < 0 or new_row >= N: continue
        if new_col < 0 or new_col >= N: continue
        if board[row][col] == board[new_row][new_col]:
            union(normal_parent, ((N * row) + col), ((N * new_row) + new_col))
        if board[row][col] in ("R", "G") and board[new_row][new_col] in ("R", "G") or board[row][col] == board[new_row][new_col] == "B":
            union(color_blindness_parent, (N * row + col), (N * new_row + new_col))
print(len(Counter(normal_parent).keys()), len(Counter(color_blindness_parent).keys()))

결국 유니온 파인드 포기하고 BFS로 틀었씁니다..

from collections import deque

import sys

def input(): return sys.stdin.readline().rstrip()

N = int(input())

idx = 0

board = []

for _ in range(N):
board.append(list(input()))
dx = [0, 0, -1, 1]

dy = [-1, 1, 0, 0]

visited = set()

deq = deque()

normal_answer = 0

for row in range(N):
for col in range(N):
    if (row, col) in visited: continue
    normal_answer += 1
    visited.add((row, col))
    deq.append((row, col))
    while deq:
        now_row, now_col = deq.popleft()
        for dir in range(4):
            new_row = now_row + dy[dir]
            new_col = now_col + dx[dir]
            if new_row < 0 or new_row >= N: continue
            if new_col < 0 or new_col >= N: continue
            if (new_row, new_col) in visited: continue
            if board[row][col] == board[new_row][new_col]:
                deq.append((new_row, new_col))
                visited.add((new_row, new_col))
visited = set()

deq = deque()

color_blindness_answer = 0

for row in range(N):
for col in range(N):
    if (row, col) in visited: continue
    color_blindness_answer += 1
    visited.add((row, col))
    deq.append((row, col))
    while deq:
        now_row, now_col = deq.popleft()
        for dir in range(4):
            new_row = now_row + dy[dir]
            new_col = now_col + dx[dir]
            if new_row < 0 or new_row >= N: continue
            if new_col < 0 or new_col >= N: continue
            if (new_row, new_col) in visited: continue
            if board[row][col] in ("R", "G") and board[new_row][new_col] in ("R", "G") or board[row][col] == board[new_row][new_col] == "B":
                deq.append((new_row, new_col))
                visited.add((new_row, new_col))
print(normal_answer, color_blindness_answer)

와 근데 R이랑 G 똑같이 맞추고 다시 돌리는 거 뤠전두다 뤠전두

유니온파인드 코드 반례 찾았습니다
20

BBBBBRRRRRRRRRRRBBBB

BBBBBRRRRRRRRRRRBBBB

RBBBBBRRRRRRRRRRBBBB

RRRBBBBRRRRRRRRRBBBB

RRRBBBBRRRRRRRRRRBRB

GRRBBBBRRRRRRRRRRBRR

GGRRRRBBBRRRRRRRRBBB

GGGRRRBBBRRRRRRRRBBB

RRGGGGBBBRRRRRRRRBBB

BBGGGGBBBBRRRRRRRBBB

BBGGGGGBBBRRRRRRRBBB

GBGGGGGBRRRRRRRRRBBB

GGGGGGGGRRRRRRRRRBBB

GGGGGGGGGRRRRRRRRBBB

GGGGGGGGGGRRRRRRRBBB

RRGGGGGGGGGGRRRRRRBB

RRGGGGGGGGGGGGGGGRBB

RRRGGGGGGGGGGGGGGRBB

GGGGGGGBGGGGGGGGGGBB

RRRRGGGGGGGGGGGGGGGG
를 입력하면 답은 11 6인데 12 7이 나오네용
하 근데 왜 안되는지는 저도 아직 못찾겠네요.. 쫌 더 생각해보겠슴둥

@tgyuuAn 찾았습니다 ㅎㅎ
for i in range(len(normal_parent)):

    print(i, " : ", normal_parent[i])
를 돌리면 아래처럼 나와요

그런데 실제로 361번째 값을 보면 부모가 100이어야 하는데 부모가 360으로 되어 있어요(0~400번기준으로 index를 잡은 경우입니다)
그래서

print 출력해서 돌려보니까
graph[360] 값이 100으로 변경되기 전에 이미 graph[361]을 360으로 바꿨기 때문에 graph[361] 100으로 업데이트가 안된 것 같아요!!
그래서
최종 print 문을 출력하기 전에, 한번더 find_parent()를 돌려서 부모노드를 가르키도록 다시 업데이트 해주어야 할 것 같습니다....
애렵네요
유니온 파인드가 원래 이런 경우를 다 생각해주어야 할까요................???
대박적이네뇨.. 와퐁쥬 대박...

생각해보니 마지막에 find_parent() 돌려주어서 변경된 진짜 부모를 찾아줘야해요

저렇게 하묜 시간안에 세잎 되나요?!?!?

예압 통과됩니당 ~~

2024-05-22

43ee014

H0ngJu added 작성 중 ⏱️ H0ngJu labels May 22, 2024

H0ngJu self-assigned this May 22, 2024

H0ngJu added 리뷰 기다리는 중 🔥 and removed 작성 중 ⏱️ labels May 22, 2024

github-actions bot requested review from alstjr7437, SeongHoonC and tgyuuAn May 22, 2024 15:49

SeongHoonC approved these changes May 23, 2024

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17-H0ngJu #195

17-H0ngJu #195

H0ngJu commented May 22, 2024 •

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H0ngJu commented May 26, 2024 •

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tgyuuAn commented May 27, 2024 •

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H0ngJu commented May 27, 2024

를 입력하면 답은 11 6인데 12 7이 나오네용
하 근데 왜 안되는지는 저도 아직 못찾겠네요.. 쫌 더 생각해보겠슴둥

17-H0ngJu #195

17-H0ngJu #195

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H0ngJu commented May 22, 2024 • edited Loading

🔗 문제 링크

✔️ 소요된 시간

✨ 수도 코드

📚 새롭게 알게된 내용

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H0ngJu commented May 27, 2024

를 입력하면 답은 11 6인데 12 7이 나오네용 하 근데 왜 안되는지는 저도 아직 못찾겠네요.. 쫌 더 생각해보겠슴둥

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tgyuuAn commented May 27, 2024 •

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를 입력하면 답은 11 6인데 12 7이 나오네용
하 근데 왜 안되는지는 저도 아직 못찾겠네요.. 쫌 더 생각해보겠슴둥