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10.2 &nbsp; 二分查找插入点
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10.3 &nbsp; 二分查找边界
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第 12 章 &nbsp; 分治
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12.1 &nbsp; 分治算法
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12.2 &nbsp; 分治搜索策略
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12.3 &nbsp; 构建树问题
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12.4 &nbsp; 汉诺塔问题
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12.5 &nbsp; 小结
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第 14 章 &nbsp; 动态规划
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14.1 &nbsp; 初探动态规划
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14.2 &nbsp; DP 问题特性
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14.3 &nbsp; DP 解题思路
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14.4 &nbsp; 0-1 背包问题
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14.4 &nbsp; 0-1 背包问题
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14.5 &nbsp; 完全背包问题
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14.6 &nbsp; 编辑距离问题
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14.7 &nbsp; 小结
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第 15 章 &nbsp; 贪心
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15.1 &nbsp; 贪心算法
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15.2 &nbsp; 分数背包问题
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15.3 &nbsp; 最大容量问题
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15.4 &nbsp; 最大切分乘积问题
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15.5 &nbsp; 小结
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<p>给定 <span class="arithmatex">\(n\)</span> 个物品，第 <span class="arithmatex">\(i\)</span> 个物品的重量为 <span class="arithmatex">\(wgt[i-1]\)</span>、价值为 <span class="arithmatex">\(val[i-1]\)</span> ，和一个容量为 <span class="arithmatex">\(cap\)</span> 的背包。每个物品只能选择一次，问在不超过背包容量下能放入物品的最大价值。</p>
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<p>观察图 14-17 ，由于物品编号 <span class="arithmatex">\(i\)</span> 从 <span class="arithmatex">\(1\)</span> 开始计数，数组索引从 <span class="arithmatex">\(0\)</span> 开始计数，因此物品 <span class="arithmatex">\(i\)</span> 对应重量 <span class="arithmatex">\(wgt[i-1]\)</span> 和价值 <span class="arithmatex">\(val[i-1]\)</span> 。</p>
<p><img alt="0-1 背包的示例数据" src="../knapsack_problem.assets/knapsack_example.png" /></p>
<p><a class="glightbox" href="../knapsack_problem.assets/knapsack_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="0-1 背包的示例数据" src="../knapsack_problem.assets/knapsack_example.png" /></a></p>
<p align="center"> 图 14-17 &nbsp; 0-1 背包的示例数据 </p>
<p>我们可以将 0-1 背包问题看作是一个由 <span class="arithmatex">\(n\)</span> 轮决策组成的过程，每个物体都有不放入和放入两种决策，因此该问题是满足决策树模型的。</p>
@ -3770,7 +3597,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
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<p>如图 14-18 所示，由于每个物品都会产生不选和选两条搜索分支，因此时间复杂度为 <span class="arithmatex">\(O(2^n)\)</span> 。</p>
<p>观察递归树，容易发现其中存在重叠子问题，例如 <span class="arithmatex">\(dp[1, 10]\)</span> 等。而当物品较多、背包容量较大，尤其是相同重量的物品较多时，重叠子问题的数量将会大幅增多。</p>
<p><img alt="0-1 背包的暴力搜索递归树" src="../knapsack_problem.assets/knapsack_dfs.png" /></p>
<p><a class="glightbox" href="../knapsack_problem.assets/knapsack_dfs.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="0-1 背包的暴力搜索递归树" src="../knapsack_problem.assets/knapsack_dfs.png" /></a></p>
<p align="center"> 图 14-18 &nbsp; 0-1 背包的暴力搜索递归树 </p>
<h3 id="2">2. &nbsp; 方法二：记忆化搜索<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
@ -4061,7 +3888,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
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<p>图 14-19 展示了在记忆化递归中被剪掉的搜索分支。</p>
<p><img alt="0-1 背包的记忆化搜索递归树" src="../knapsack_problem.assets/knapsack_dfs_mem.png" /></p>
<p><a class="glightbox" href="../knapsack_problem.assets/knapsack_dfs_mem.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="0-1 背包的记忆化搜索递归树" src="../knapsack_problem.assets/knapsack_dfs_mem.png" /></a></p>
<p align="center"> 图 14-19 &nbsp; 0-1 背包的记忆化搜索递归树 </p>
<h3 id="3">3. &nbsp; 方法三：动态规划<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
@ -4333,46 +4160,46 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
<div class="tabbed-set tabbed-alternate" data-tabs="4:14"><input checked="checked" id="__tabbed_4_1" name="__tabbed_4" type="radio" /><input id="__tabbed_4_2" name="__tabbed_4" type="radio" /><input id="__tabbed_4_3" name="__tabbed_4" type="radio" /><input id="__tabbed_4_4" name="__tabbed_4" type="radio" /><input id="__tabbed_4_5" name="__tabbed_4" type="radio" /><input id="__tabbed_4_6" name="__tabbed_4" type="radio" /><input id="__tabbed_4_7" name="__tabbed_4" type="radio" /><input id="__tabbed_4_8" name="__tabbed_4" type="radio" /><input id="__tabbed_4_9" name="__tabbed_4" type="radio" /><input id="__tabbed_4_10" name="__tabbed_4" type="radio" /><input id="__tabbed_4_11" name="__tabbed_4" type="radio" /><input id="__tabbed_4_12" name="__tabbed_4" type="radio" /><input id="__tabbed_4_13" name="__tabbed_4" type="radio" /><input id="__tabbed_4_14" name="__tabbed_4" type="radio" /><div class="tabbed-labels"><label for="__tabbed_4_1">&lt;1&gt;</label><label for="__tabbed_4_2">&lt;2&gt;</label><label for="__tabbed_4_3">&lt;3&gt;</label><label for="__tabbed_4_4">&lt;4&gt;</label><label for="__tabbed_4_5">&lt;5&gt;</label><label for="__tabbed_4_6">&lt;6&gt;</label><label for="__tabbed_4_7">&lt;7&gt;</label><label for="__tabbed_4_8">&lt;8&gt;</label><label for="__tabbed_4_9">&lt;9&gt;</label><label for="__tabbed_4_10">&lt;10&gt;</label><label for="__tabbed_4_11">&lt;11&gt;</label><label for="__tabbed_4_12">&lt;12&gt;</label><label for="__tabbed_4_13">&lt;13&gt;</label><label for="__tabbed_4_14">&lt;14&gt;</label></div>
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<p><img alt="0-1 背包的动态规划过程" src="../knapsack_problem.assets/knapsack_dp_step1.png" /></p>
<p><a class="glightbox" href="../knapsack_problem.assets/knapsack_dp_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="0-1 背包的动态规划过程" src="../knapsack_problem.assets/knapsack_dp_step1.png" /></a></p>
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@ -4389,22 +4216,22 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
<div class="tabbed-set tabbed-alternate" data-tabs="5:6"><input checked="checked" id="__tabbed_5_1" name="__tabbed_5" type="radio" /><input id="__tabbed_5_2" name="__tabbed_5" type="radio" /><input id="__tabbed_5_3" name="__tabbed_5" type="radio" /><input id="__tabbed_5_4" name="__tabbed_5" type="radio" /><input id="__tabbed_5_5" name="__tabbed_5" type="radio" /><input id="__tabbed_5_6" name="__tabbed_5" type="radio" /><div class="tabbed-labels"><label for="__tabbed_5_1">&lt;1&gt;</label><label for="__tabbed_5_2">&lt;2&gt;</label><label for="__tabbed_5_3">&lt;3&gt;</label><label for="__tabbed_5_4">&lt;4&gt;</label><label for="__tabbed_5_5">&lt;5&gt;</label><label for="__tabbed_5_6">&lt;6&gt;</label></div>
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<p><img alt="0-1 背包的空间优化后的动态规划过程" src="../knapsack_problem.assets/knapsack_dp_comp_step1.png" /></p>
<p><a class="glightbox" href="../knapsack_problem.assets/knapsack_dp_comp_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="0-1 背包的空间优化后的动态规划过程" src="../knapsack_problem.assets/knapsack_dp_comp_step1.png" /></a></p>
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