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chapter_dynamic_programming/dp_problem_features/index.html

@@ -3407,7 +3407,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
 <p>本题显然具有最优子结构：我们从两个子问题最优解 <span class="arithmatex">\(dp[i-1]\)</span> , <span class="arithmatex">\(dp[i-2]\)</span> 中挑选出较优的那一个，并用它构建出原问题 <span class="arithmatex">\(dp[i]\)</span> 的最优解。</p>
 <p>那么，上节的爬楼梯题目有没有最优子结构呢？它的目标是求解方案数量，看似是一个计数问题，但如果换一种问法：“求解最大方案数量”。我们意外地发现，<strong>虽然题目修改前后是等价的，但最优子结构浮现出来了</strong>：第 <span class="arithmatex">\(n\)</span> 阶最大方案数量等于第 <span class="arithmatex">\(n-1\)</span> 阶和第 <span class="arithmatex">\(n-2\)</span> 阶最大方案数量之和。所以说，最优子结构的解释方式比较灵活，在不同问题中会有不同的含义。</p>
 <p>根据状态转移方程，以及初始状态 <span class="arithmatex">\(dp[1] = cost[1]\)</span> , <span class="arithmatex">\(dp[2] = cost[2]\)</span> ，可以得出动态规划代码。</p>
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+<div class="tabbed-set tabbed-alternate" data-tabs="1:12"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><input id="__tabbed_1_6" name="__tabbed_1" type="radio" /><input id="__tabbed_1_7" name="__tabbed_1" type="radio" /><input id="__tabbed_1_8" name="__tabbed_1" type="radio" /><input id="__tabbed_1_9" name="__tabbed_1" type="radio" /><input id="__tabbed_1_10" name="__tabbed_1" type="radio" /><input id="__tabbed_1_11" name="__tabbed_1" type="radio" /><input id="__tabbed_1_12" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Java</label><label for="__tabbed_1_2">C++</label><label for="__tabbed_1_3">Python</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">JS</label><label for="__tabbed_1_6">TS</label><label for="__tabbed_1_7">C</label><label for="__tabbed_1_8">C#</label><label for="__tabbed_1_9">Swift</label><label for="__tabbed_1_10">Zig</label><label for="__tabbed_1_11">Dart</label><label for="__tabbed_1_12">Rust</label></div>
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 <div class="highlight"><span class="filename">min_cost_climbing_stairs_dp.java</span><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="cm">/* 爬楼梯最小代价：动态规划 */</span>
@@ -3582,7 +3582,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
 <p align="center"> Fig. 爬楼梯最小代价的动态规划过程 </p>
 
 <p>本题也可以进行状态压缩，将一维压缩至零维，使得空间复杂度从 <span class="arithmatex">\(O(n)\)</span> 降低至 <span class="arithmatex">\(O(1)\)</span> 。</p>
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+<div class="tabbed-set tabbed-alternate" data-tabs="2:12"><input checked="checked" id="__tabbed_2_1" name="__tabbed_2" type="radio" /><input id="__tabbed_2_2" name="__tabbed_2" type="radio" /><input id="__tabbed_2_3" name="__tabbed_2" type="radio" /><input id="__tabbed_2_4" name="__tabbed_2" type="radio" /><input id="__tabbed_2_5" name="__tabbed_2" type="radio" /><input id="__tabbed_2_6" name="__tabbed_2" type="radio" /><input id="__tabbed_2_7" name="__tabbed_2" type="radio" /><input id="__tabbed_2_8" name="__tabbed_2" type="radio" /><input id="__tabbed_2_9" name="__tabbed_2" type="radio" /><input id="__tabbed_2_10" name="__tabbed_2" type="radio" /><input id="__tabbed_2_11" name="__tabbed_2" type="radio" /><input id="__tabbed_2_12" name="__tabbed_2" type="radio" /><div class="tabbed-labels"><label for="__tabbed_2_1">Java</label><label for="__tabbed_2_2">C++</label><label for="__tabbed_2_3">Python</label><label for="__tabbed_2_4">Go</label><label for="__tabbed_2_5">JS</label><label for="__tabbed_2_6">TS</label><label for="__tabbed_2_7">C</label><label for="__tabbed_2_8">C#</label><label for="__tabbed_2_9">Swift</label><label for="__tabbed_2_10">Zig</label><label for="__tabbed_2_11">Dart</label><label for="__tabbed_2_12">Rust</label></div>
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 <div class="highlight"><span class="filename">min_cost_climbing_stairs_dp.java</span><pre><span></span><code><a id="__codelineno-12-1" name="__codelineno-12-1" href="#__codelineno-12-1"></a><span class="cm">/* 爬楼梯最小代价：状态压缩后的动态规划 */</span>
@@ -3760,7 +3760,7 @@ dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
 <p align="center"> Fig. 考虑约束下的递推关系 </p>
 
 <p>最终，返回 <span class="arithmatex">\(dp[n, 1] + dp[n, 2]\)</span> 即可，两者之和代表爬到第 <span class="arithmatex">\(n\)</span> 阶的方案总数。</p>
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 <div class="highlight"><span class="filename">climbing_stairs_constraint_dp.java</span><pre><span></span><code><a id="__codelineno-24-1" name="__codelineno-24-1" href="#__codelineno-24-1"></a><span class="cm">/* 带约束爬楼梯：动态规划 */</span>