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chapter_backtracking/backtracking_algorithm/index.html
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Copyright &copy; 2022-2024 krahets<br>The website content is licensed under <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a>
Copyright &copy; 2024 krahets<br>The website content is licensed under <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a>
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chapter_backtracking/index.html
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Copyright &copy; 2022-2024 krahets<br>The website content is licensed under <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a>
Copyright &copy; 2024 krahets<br>The website content is licensed under <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a>
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chapter_backtracking/n_queens_problem/index.html
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<h3 id="1">1. &nbsp; 逐行放置策略<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>皇后的数量和棋盘的行数都为 <span class="arithmatex">\(n\)</span> ，因此我们容易得到一个推论：<strong>棋盘每行都允许且只允许放置一个皇后</strong>。</p>
<p>也就是说，我们可以采取逐行放置策略：从第一行开始，在每行放置一个皇后，直至最后一行结束。</p>
<p>图 13-17 所示为 <span class="arithmatex">\(4\)</span> 皇后问题的逐行放置过程。受画幅限制，图 13-17 仅展开了第一行的其中一个搜索分支，并且将不满足列约束和对角线约束的方案都进行了剪枝。</p>
<p>图 13-17 所示为 4 皇后问题的逐行放置过程。受画幅限制，图 13-17 仅展开了第一行的其中一个搜索分支，并且将不满足列约束和对角线约束的方案都进行了剪枝。</p>
<p><a class="glightbox" href="../n_queens_problem.assets/n_queens_placing.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="逐行放置策略" class="animation-figure" src="../n_queens_problem.assets/n_queens_placing.png" /></a></p>
<p align="center"> 图 13-17 &nbsp; 逐行放置策略 </p>
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Copyright &copy; 2022-2024 krahets<br>The website content is licensed under <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a>
Copyright &copy; 2024 krahets<br>The website content is licensed under <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a>
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chapter_backtracking/permutations_problem/index.html
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Copyright &copy; 2022-2024 krahets<br>The website content is licensed under <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a>
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chapter_backtracking/subset_sum_problem/index.html
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Copyright &copy; 2022-2024 krahets<br>The website content is licensed under <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a>
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chapter_backtracking/summary/index.html
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Copyright &copy; 2022-2024 krahets<br>The website content is licensed under <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a>
Copyright &copy; 2024 krahets<br>The website content is licensed under <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/">CC BY-NC-SA 4.0</a>
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