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jaeseo222
reviewed
Nov 29, 2022
Comment on lines
+68
to
+72
| //진실을 아는 사람은 한 집합에 | ||
| while (k--) { | ||
| cin >> h; | ||
| unionInput(base, h); | ||
| } |
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p3. 진실을 아는 사람들을 한 집합에 연결해야 한다는 걸 잘 캐치 해주셨네요!!✨✨✨
그런데, 현재 작성해주신 코드로는 한 집합의 부모가 입력으로 들어오는 사람의 번호 중 하나네요!
이렇게 되면 어떤 사람의 번호가 부모인지 예제마다 달라지겠네요!
달라지지 않기 위해선, 우리가 임의의 번호를 주어주는 건 어떨까요? 입력으로 들어오지 않는 사람의 번호로요!
Comment on lines
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| int countStory(vector<vector<int>>& party, int base) { | ||
| int cnt = 0; | ||
| bool flag = true; | ||
| for (int i = 0; i < party.size(); i++) { | ||
| flag = true; | ||
| for (int j = 0; j < party[i].size(); j++) { | ||
| //각 파티를 다시 체크하면서 파티에 진실을 아는 집합의 사람이 한 명이라도 있으면 | ||
| //cnt를 더하지 않음 | ||
| if (parent[base] == party[i][j]) { | ||
| flag = false; | ||
| break; | ||
| } | ||
| if (findParent(party[i][j]) == findParent(base) || parent[party[i][j]] == base) { | ||
| flag = false; | ||
| break; | ||
| } | ||
| } | ||
| if (!flag) { | ||
| continue; | ||
| } | ||
| cnt++; | ||
| } | ||
| return cnt; | ||
| } |
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p2. 진실을 아는 사람들을 임의의 한 집합에 연결하게 되면, 임의의 숫자를 알고 있으니 이 부분도 간단해지겠네요!😁😁
|
P1. 안녕하세요 영은님! |
Dong-droid
reviewed
Dec 1, 2022
| int count_right(int row, int col, int flag) { | ||
| int cnt = 0; | ||
| for (int i = col; i < MAX; i++) { | ||
| cout << "col"; |
Dong-droid
reviewed
Dec 1, 2022
| if (board[i][j] == 0) { | ||
| continue; | ||
| } | ||
| int k = count_right(i, j, board[i][j]); |
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