[ DFS, BFS ] 10월 9일 #7
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Goldchae
approved these changes
Oct 14, 2023
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| return 0; |
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| void dfs(int c, vector<bool>& visited, vector<vector<int>>& com, int& cnt){ | ||
| visited[c]=true; | ||
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| for(int i=0;i<com[c].size();i++){ | ||
| int temp=com[c][i]; | ||
| if(!visited[temp]){ | ||
| cnt++; | ||
| dfs(temp, visited, com, cnt); | ||
| } | ||
| } | ||
| } |
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참조로 넘겨 주신 점, 깔끔한 기본 DFS 함수 좋습니다!
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이 문제는 BFS로는 쓸 수 없을까요? 가능하다면 한번 작성해보세요! DFS 함수도 스택을 이용해서 구현해보셔도 도움이 많이 될 것 같습니다!
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| cout << count << '\n'; |
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| vector<int> dir_x={-1, 1, -1, 1, 0, 0, -1, 1}; | ||
| vector<int> dir_y={-1, 1, 1, -1, 1, -1, 0, 0}; | ||
| int board[50][50]; |
| return (x>=0 && x<h && y>=0 && y<w); | ||
| } | ||
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| void dfs(int x, int y, int w, int h){ |
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이 문제도 BFS로는 쓸 수 없을까요? 가능하다면 한번 작성해보세요!
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