[김소희] 11주차 과제 제출합니다. #18
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kth990303
requested changes
May 22, 2022
| void dfs(int cnt, int idx) { //������ ��� ��, ���� ��� | ||
| if (cnt >= result) return; //���� ����� ������ ��� ���� ����� �ʿ��� ��� return | ||
| if (cnt > 0) { | ||
| bool solved[11] = { false, }; |
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,없어도 모두 초기화됩니다
Suggested change
| bool solved[11] = { false, }; | |
| bool solved[11] = { false }; |
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| for (int i = 0; i < cnt; i++) { | ||
| int cidx = team[i]; | ||
| //���� ����� Ǯ �� �ִ� ��� �˻� | ||
| for (int j = 0; j < can_solve[cidx].size(); j++) | ||
| solved[can_solve[cidx][j]] = true; | ||
| } | ||
| bool avail = true; | ||
| for (int i = 1; i <= problem; i++) { | ||
| if (!solved[i]) { | ||
| avail = false; | ||
| break; | ||
| } | ||
| } | ||
| //��� ����� Ǯ �� �ִٸ� �ּ� ��� ���� | ||
| if (avail) | ||
| result = min(result, cnt); | ||
| } |
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Challenging 🆙
정말 놀랍게도 이 작업을 안할 수 있는 방법이 존재합니다. 하지만 지금 이 방법을 생각해내기엔 어려울 수 있겠네요
| can_solve[i].push_back(num); | ||
| } | ||
| } | ||
| dfs(0, 1); |
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소희님에게 이 문제가 어떤 부분에서 코드 작성이 어려웠는지 궁금합니다.
어디서 막히셨는지 한 번 설명해주시면 감사하겠습니다!
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| if (result == 11) result = -1; | ||
| cout << result; | ||
| return 0; |
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삼항연산자 이용하는 방법도 존재합니다.
Suggested change
| if (result == 11) result = -1; | |
| cout << result; | |
| return 0; | |
| cout << (result==11 ? -1 : result); |
| { | ||
| cout << blue.at(i) << " "; | ||
| } | ||
| cout << endl << white.size() << endl; |
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'\n' 쓰기 힘들거나 귀찮을 때
#define endl '\n'
해놓으면 편리합니다.
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| if (it1 == blue.end()&&sum==0) { | ||
| blue.push_back(i); | ||
| sum++; | ||
| } | ||
| if (it2 == white.end()&&sum==0) { | ||
| white.push_back(i); | ||
| sum++; | ||
| } | ||
| if(sum==0) | ||
| blue.push_back(i); |
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싫어하는 사람이 없는 사람이 일부 존재한다면 어떻게 될까요?
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| if (white.size() == 0) { | ||
| blue.erase(blue.begin() + 1); | ||
| white.push_back(1); | ||
| }if (blue.size() == 0) { | ||
| white.erase(white.begin() + 1); | ||
| blue.push_back(1); | ||
| } |
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계속 생각해보고 있는데 너무 어렵네요ㅠㅠ