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第 8 章 &nbsp; 堆積
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第 9 章 &nbsp;
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第 9 章 &nbsp;
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9.2 &nbsp; 圖基礎操作
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第 10 章 &nbsp; 搜尋
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10.2 &nbsp; 二分搜尋插入點
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10.3 &nbsp; 二分搜尋邊界
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10.4 &nbsp; 雜湊最佳化策略
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10.5 &nbsp; 重識搜尋演算法
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第 11 章 &nbsp; 排序
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11.2 &nbsp; 選擇排序
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第 12 章 &nbsp; 分治
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12.1 &nbsp; 分治演算法
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第 13 章 &nbsp; 回溯
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<h1 id="147">14.7 &nbsp; 小結<a class="headerlink" href="#147" title="Permanent link">&para;</a></h1>
<ul>
<li>動態規劃對問題進行分解,並透過儲存子問題的解來規避重複計算,提高計算效率。</li>
<li>不考慮時間的前提下,所有動態規劃問題都可以用回溯(暴力搜尋)進行求解,但遞迴樹中存在大量的重疊子問題,效率極低。透過引入記憶化串列,可以儲存所有計算過的子問題的解,從而保證重疊子問題只被計算一次。</li>
<li>記憶化搜尋是一種從頂至底的遞迴式解法,而與之對應的動態規劃是一種從底至頂的遞推式解法,其如同“填寫表格”一樣。由於當前狀態僅依賴某些區域性狀態,因此我們可以消除 <span class="arithmatex">\(dp\)</span> 表的一個維度,從而降低空間複雜度。</li>
<li>子問題分解是一種通用的演算法思路,在分治、動態規劃、回溯中具有不同的性質。</li>
<li>動態規劃問題有三大特性:重疊子問題、最優子結構、無後效性。</li>
<li>如果原問題的最優解可以從子問題的最優解構建得來,則它就具有最優子結構。</li>
<li>無後效性指對於一個狀態,其未來發展只與該狀態有關,而與過去經歷的所有狀態無關。許多組合最佳化問題不具有無後效性,無法使用動態規劃快速求解。</li>
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<p><strong>背包問題</strong></p>
<ul>
<li>背包問題是最典型的動態規劃問題之一,具有 0-1 背包、完全背包、多重背包等變種。</li>
<li>0-1 背包的狀態定義為前 <span class="arithmatex">\(i\)</span> 個物品在容量為 <span class="arithmatex">\(c\)</span> 的背包中的最大價值。根據不放入背包和放入背包兩種決策,可得到最優子結構,並構建出狀態轉移方程。在空間最佳化中,由於每個狀態依賴正上方和左上方的狀態,因此需要倒序走訪串列,避免左上方狀態被覆蓋。</li>
<li>完全背包問題的每種物品的選取數量無限制,因此選擇放入物品的狀態轉移與 0-1 背包問題不同。由於狀態依賴正上方和正左方的狀態,因此在空間最佳化中應當正序走訪。</li>
<li>零錢兌換問題是完全背包問題的一個變種。它從求“最大”價值變為求“最小”硬幣數量,因此狀態轉移方程中的 <span class="arithmatex">\(\max()\)</span> 應改為 <span class="arithmatex">\(\min()\)</span> 。從追求“不超過”背包容量到追求“恰好”湊出目標金額,因此使用 <span class="arithmatex">\(amt + 1\)</span> 來表示“無法湊出目標金額”的無效解。</li>
<li>零錢兌換問題 II 從求“最少硬幣數量”改為求“硬幣組合數量”,狀態轉移方程相應地從 <span class="arithmatex">\(\min()\)</span> 改為求和運算子。</li>
</ul>
<p><strong>編輯距離問題</strong></p>
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<li>編輯距離Levenshtein 距離)用於衡量兩個字串之間的相似度,其定義為從一個字串到另一個字串的最少編輯步數,編輯操作包括新增、刪除、替換。</li>
<li>編輯距離問題的狀態定義為將 <span class="arithmatex">\(s\)</span> 的前 <span class="arithmatex">\(i\)</span> 個字元更改為 <span class="arithmatex">\(t\)</span> 的前 <span class="arithmatex">\(j\)</span> 個字元所需的最少編輯步數。當 <span class="arithmatex">\(s[i] \ne t[j]\)</span> 時,具有三種決策:新增、刪除、替換,它們都有相應的剩餘子問題。據此便可以找出最優子結構與構建狀態轉移方程。而當 <span class="arithmatex">\(s[i] = t[j]\)</span> 時,無須編輯當前字元。</li>
<li>在編輯距離中,狀態依賴其正上方、正左方、左上方的狀態,因此空間最佳化後正序或倒序走訪都無法正確地進行狀態轉移。為此,我們利用一個變數暫存左上方狀態,從而轉化到與完全背包問題等價的情況,可以在空間最佳化後進行正序走訪。</li>
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