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<h1 id="137">13.7. &nbsp; 小结<a class="headerlink" href="#137" title="Permanent link">&para;</a></h1>
<ul>
<li>动态规划通过将原问题分解为子问题来求解问题,并通过存储子问题的解来规避重复计算,实现高效的计算效率。子问题分解是一种通用的算法思路,在分治、动态规划、回溯中具有不同的性质。</li>
<li>不考虑时间的前提下,所有动态规划问题都可以用回溯(暴力搜索)进行求解,但递归树中存在大量的重叠子问题,效率极低。通过引入记忆化列表,可以存储所有计算过的子问题的解,从而保证重叠子问题只被计算一次。</li>
<li>记忆化递归是一种从顶至底的递归式解法,而与之对应的动态规划是一种从底至顶的递推式解法,就像是在“填写表格”一样。由于当前状态仅依赖于某些局部状态,因此我们可以消除 <span class="arithmatex">\(dp\)</span> 表的一个维度,从而降低空间复杂度。</li>
<li>动态规划问题的三大特性:重叠子问题、最优子结构、无后效性。如果原问题的最优解可以从子问题的最优解构建得来,则此问题就具有最优子结构。无后效性指对于一个状态,其未来发展只与该状态有关,与其所经历的过去的所有状态无关。许多组合优化问题都不具有无后效性,无法使用动态规划快速求解。</li>
<li>背包问题是最典型的动态规划题目,具有 0-1 背包、完全背包、多重背包等变种问题。</li>
<li>0-1 背包的状态定义为前 <span class="arithmatex">\(i\)</span> 个物品在剩余容量为 <span class="arithmatex">\(c\)</span> 的背包中的最大价值。这是一种常见的定义方式。不放入物品 <span class="arithmatex">\(i\)</span> ,状态转移至 <span class="arithmatex">\([i-1, c]\)</span> ,放入则转移至 <span class="arithmatex">\([i-1, c-wgt[i-1]]\)</span> ,由此便得到最优子结构,并构建出状态转移方程。对于状态压缩,由于每个状态依赖正上方和左上方的状态,因此需要倒序遍历列表,避免左上方状态被覆盖。</li>
<li>完全背包的每种物品有无数个,因此在放置物品 <span class="arithmatex">\(i\)</span> 后,状态转移至 <span class="arithmatex">\([i, c-wgt[i-1]]\)</span> 。由于状态依赖于正上方和正左方的状态,因此状态压缩后应该正序遍历。</li>
<li>零钱兑换问题是完全背包的一个变种。为从求“最大“价值变为求“最小”硬币数量,我们将状态转移方程中的 <span class="arithmatex">\(\max()\)</span> 改为 <span class="arithmatex">\(\min()\)</span> 。为从求“不超过”背包容量到求“恰好”凑出目标金额,我们使用 <span class="arithmatex">\(amt + 1\)</span> 来表示“无法凑出目标金额”的无效解。</li>
<li>零钱兑换 II 问题从求“最少硬币数量”改为求“硬币组合数量”,状态转移方程相应地从 <span class="arithmatex">\(\min()\)</span> 改为求和运算符。</li>
<li>编辑距离Levenshtein 距离)用于衡量两个字符串之间的相似度,定义为从一个字符串到另一个字符串的最小编辑步数,编辑操作包括添加、删除、替换。</li>
<li>编辑距离问题的状态定义为将 <span class="arithmatex">\(s\)</span> 的前 <span class="arithmatex">\(i\)</span> 个字符更改为 <span class="arithmatex">\(t\)</span> 的前 <span class="arithmatex">\(j\)</span> 个字符所需的最少编辑步数。考虑字符 <span class="arithmatex">\(s[i]\)</span><span class="arithmatex">\(t[j]\)</span> ,具有三种决策:在 <span class="arithmatex">\(s[i-1]\)</span> 之后添加 <span class="arithmatex">\(t[j-1]\)</span> 、删除 <span class="arithmatex">\(s[i-1]\)</span> 、将 <span class="arithmatex">\(s[i-1]\)</span> 替换为 <span class="arithmatex">\(t[j-1]\)</span> ,它们都有相应的剩余子问题,据此就可以找出最优子结构与构建状态转移方程。值得注意的是,当 <span class="arithmatex">\(s[i] = t[j]\)</span> 时,无需编辑当前字符,直接跳过即可。</li>
<li>在编辑距离中,状态依赖于其正上方、正左方、左上方的状态,因此状态压缩后正序或倒序遍历都无法正确地进行状态转移。利用一个变量暂存左上方状态,即转化至完全背包地情况,可以在状态压缩后使用正序遍历。</li>
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