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krahets
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<span class="md-ellipsis">
8.3 &nbsp; Top-K 问题
8.3 &nbsp; Top-k 问题
</span>
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<p><a class="glightbox" href="../dp_problem_features.assets/climbing_stairs_constraint_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="带约束爬到第 3 阶的方案数量" class="animation-figure" src="../dp_problem_features.assets/climbing_stairs_constraint_example.png" /></a></p>
<p align="center"> 图 14-8 &nbsp; 带约束爬到第 3 阶的方案数量 </p>
<p>在该问题中,如果上一轮是跳 <span class="arithmatex">\(1\)</span> 阶上来的,那么下一轮就必须跳 <span class="arithmatex">\(2\)</span> 阶。这意味着,<strong>下一步选择不能由当前状态(当前所在楼梯阶数)独立决定,还和前一个状态(上轮所在楼梯阶数)有关</strong></p>
<p>在该问题中,如果上一轮是跳 <span class="arithmatex">\(1\)</span> 阶上来的,那么下一轮就必须跳 <span class="arithmatex">\(2\)</span> 阶。这意味着,<strong>下一步选择不能由当前状态(当前所在楼梯阶数)独立决定,还和前一个状态(上轮所在楼梯阶数)有关</strong></p>
<p>不难发现,此问题已不满足无后效性,状态转移方程 <span class="arithmatex">\(dp[i] = dp[i-1] + dp[i-2]\)</span> 也失效了,因为 <span class="arithmatex">\(dp[i-1]\)</span> 代表本轮跳 <span class="arithmatex">\(1\)</span> 阶,但其中包含了许多“上一轮是跳 <span class="arithmatex">\(1\)</span> 阶上来的”方案,而为了满足约束,我们就不能将 <span class="arithmatex">\(dp[i-1]\)</span> 直接计入 <span class="arithmatex">\(dp[i]\)</span> 中。</p>
<p>为此,我们需要扩展状态定义:<strong>状态 <span class="arithmatex">\([i, j]\)</span> 表示处在第 <span class="arithmatex">\(i\)</span> 阶并且上一轮跳了 <span class="arithmatex">\(j\)</span></strong>,其中 <span class="arithmatex">\(j \in \{1, 2\}\)</span> 。此状态定义有效地区分了上一轮跳了 <span class="arithmatex">\(1\)</span> 阶还是 <span class="arithmatex">\(2\)</span> 阶,我们可以据此判断当前状态是从何而来的。</p>
<ul>