========================================= [질문 템플릿] 1. 강의 내용과 관련된 질문인가요? (예) 2. 인프런의 질문 게시판과 자주 하는 질문에 없는 내용인가요? (예) 3. 질문 잘하기 메뉴얼을 읽어보셨나요? (예) [질문 내용] 안녕하세요. 궁금증이 생겨서 질문을 드립니다. 'case문 - 문제와 풀이 - 1번 문제'에서 영한님은 사진 위쪽 부분 case 문 옆에 category 를 써서 단순 case문으로 해결 하였는데요. 여기서 저는 사진 아래쪽 부분에 적힌대로 category = '전자기기' 를 써서 해결하였습니다. 여기서 질문 입니다. category = '전자기기' 이 코드 처럼 해도, 상관 없는지 알고 싶습니다. + 아래 처럼 한 이유는 위에 쿼리문은 완전히 생각이 안났습니다.
이제 시험도 약 10일 남았는데 지금 3유형 마지막에 회귀 분석만 들으면 일단 이론이 끝나고 기출유형 공부하고자 하는데 고심끝에 8회 ~ 11회 기출에 집중하는 식으로 마무리 하고자 해요. 그 문제 2~3회 회독하면 시간 다 될꺼 같은데 어떤 식으로 하면 되는지 마지막 조언을 부탁합니다.
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안녕하세요. 빅이시 강의 수강 및 코드 관련 문의 드립니다. '추천' 파트와 달리 '기초'에는 랜덤 포레스트 모델로(만) 예측하는 코드가 구현되어 있음을 확인했습니다. 기초 케이스에도 LightGBM으로도 예측하는 코드를 추가 가능하실까요? 추천(1~3) 코드에서 LightGBM 부분을 추출하여 기초 코드를 '# 2_1. 머신러닝 학습 및 예측(랜덤포레스트)' '# 2_2. 머신러닝 학습 및 예측(LightGBM)' '# 3_1. 결과 파일 생성(랜덤포레스트)' '# 3_2. 결과 파일 생성(LightGBM)' 이렇게 코드를 자체적으로 이식(?)하고 이원화하여 수정해보려고 했는데 재대로 한 게 맞는지 확신이 없어 문의드립니다ㅠㅜ
- 학습 관련 질문을 남겨주세요. 상세히 작성하면 더 좋아요! - 먼저 유사한 질문이 있었는지 검색해보세요. - 서로 예의를 지키며 존중하는 문화를 만들어가요. - 잠깐! 인프런 서비스 운영 관련 문의는 1:1 문의하기를 이용해주세요. 안녕하세요 강사님, 저는 이 문제를 풀며 불의 최소 거리와 사람의 최소 거리를 비교하며 풀 생각을 못했고, 그래서 DFS를 사용하면서, 불을 먼저 이동시키고 사람을 이동시키면서 불을 만나지 않고 탈출하는 경우에 대해 값을 찾았었습니다. #include <bits/stdc++.h> using namespace std; int N,M; int dy[4] = {0,0,1,-1}; int dx[4] = {1,-1,0,0}; int min_exit = INT_MAX; vector<vector<int>> inp_map; vector<vector<int>> fires_map; vector<vector<pair<int,int>>> fires; vector<vector<int>> visited; bool can_exit = false; void fire_flood(int level) { for (int f=0; f<fires[level].size(); f++) { int y = fires[level][f].first; int x = fires[level][f].second; fires_map[y][x] = 1; for (int i=0; i<4; i++) { int ny = y + dy[i]; int nx = x + dx[i]; if (ny >= 0 && nx >= 0 && ny < N && nx < M && inp_map[ny][nx] != 0 && inp_map[ny][nx] != 2) { fires_map[ny][nx] = 1; fires[level+1].push_back(make_pair(ny,nx)); } } } } void return_flood(int level) { for (int f=0; f<fires[level].size(); f++) { fires_map[fires[level][f].first][fires[level][f].second] = 0; } } void dfs(int y, int x, int level) { if (y == N-1 || x == M-1 || y == 0 || x == 0) { can_exit = true; min_exit = min(min_exit, level); return; } visited[y][x] = 1; fires.push_back(vector<pair<int,int>>()); fire_flood(level); for (int i=0; i<4; i++) { int ny = y + dy[i]; int nx = x + dx[i]; if (ny >= 0 && nx >= 0 && ny < N && nx < M && inp_map[ny][nx] != 0) { if (inp_map[ny][nx] == 1 && visited[ny][nx] == 0) { dfs(ny, nx, level+1); } } } return_flood(level+1); fires.pop_back(); } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); fires.push_back(vector<pair<int,int>>()); cin >> N; cin >> M; int sty, stx; for (int i=0; i<N; i++) { inp_map.push_back(vector<int>()); fires_map.push_back(vector<int>(M,0)); visited.push_back(vector<int>(M,0)); string temp; cin >> temp; for (int j=0; j<M; j++) { if (temp[j] == '#') inp_map[i].push_back(0); else if (temp[j] == 'F') { inp_map[i].push_back(2); fires[0].push_back(make_pair(i,j)); } else if (temp[j] == 'J') { inp_map[i].push_back(3); sty = i, stx = j; } else inp_map[i].push_back(1); } } dfs(sty,stx,0); if (can_exit) cout << min_exit+1; else cout << "IMPOSSIBLE"; return 0; } 이 코드가 해당 코드인데요,, 시간복잡도에 있어서 훨씬 안좋은 코드인거같아요. 저런 불과 사람의 최단경로를 비교해야겠다! 라는 생각을 이끌어내는게 쉽지가 않은데, 어떻게 이런 생각하는 능력을 키울 수 있을까요. . .
미국 빅테크 프론트엔드 시스템 디자인 실전: 단순 구현자로 남지 않기 위한 프론트엔드 개발자를 위해
안녕하세요, 강사님. 강의 잘 보고 있습니다. Design a Toast Notification System 미션을 진행하고 있는데요, 작성한 내용을 채점하고 리뷰하기 위한 프롬프트가 있으면 좋을 것 같습니다. 강사님의 미션 평가 기준이 담긴 프롬프트로 커리큘럼을 잘 따라가고 있는지 점검하는 기회, 강사님과 수강생의 시간을 모두 아낄 수 있지 않을까 생각이 들어서 의견 남깁니다. 남은 강의도 잘 보겠습니다. 좋은 강의 제공해주셔서 감사합니다!