Site updated: 2025-03-09 14:42:34

Author: simuleite

ComputerScience/笔记/洛谷 错题本/index.html

@@ -28,7 +28,7 @@
 <meta property="og:description" content="P1004 [NOIP 2000 提高组] 方格取数 #走两次dp 如果只走一次，这题是非常经典的DP。但是要走两次，就变得非常有难度。 首先，可以简单地推广：要走两次，dp就存四个下标： int[][][][] dp &#x3D; new int[N][N][N][N]; 我们只需要遍历所有可能，并且比较四种走法（同下、同右、一下一右），取最大值就可以了。 注意，一个数只能取一次，需要一个判断防止重复取">
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 <meta property="article:published_time" content="2025-02-28T16:00:00.000Z">
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+<meta property="article:modified_time" content="2025-03-09T06:20:42.079Z">
 <meta property="article:author" content="SIMULEITE">
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@@ -141,7 +141,7 @@
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-            <div class="post-toc animated"><ol class="nav"><li class="nav-item nav-level-1"><a class="nav-link" href="#p1004-noip-2000-%E6%8F%90%E9%AB%98%E7%BB%84-%E6%96%B9%E6%A0%BC%E5%8F%96%E6%95%B0"><span class="nav-text"> P1004 [NOIP 2000 提高组] 方格取数</span></a></li><li class="nav-item nav-level-1"><a class="nav-link" href="#b3637-%E6%9C%80%E9%95%BF%E4%B8%8A%E5%8D%87%E5%AD%90%E5%BA%8F%E5%88%97"><span class="nav-text"> B3637 最长上升子序列</span></a></li><li class="nav-item nav-level-1"><a class="nav-link" href="#p2782-%E5%8F%8B%E5%A5%BD%E5%9F%8E%E5%B8%82"><span class="nav-text"> P2782 友好城市</span></a><ol class="nav-child"><li class="nav-item nav-level-2"><a class="nav-link" href="#%E8%B4%AA%E5%BF%83%E6%B3%95%E6%B1%82%E8%A7%A3lis"><span class="nav-text"> 贪心法求解LIS</span></a><ol class="nav-child"><li class="nav-item nav-level-3"><a class="nav-link" href="#%E4%BC%98%E5%8C%96dp%E6%80%9D%E8%B7%AF%E4%BA%A4%E6%8D%A2%E7%8A%B6%E6%80%81%E4%B8%8E%E7%8A%B6%E6%80%81%E5%80%BC"><span class="nav-text"> 优化DP思路：交换状态与状态值</span></a></li></ol></li></ol></li><li class="nav-item nav-level-1"><a class="nav-link" href="#p1091-noip-2004-%E6%8F%90%E9%AB%98%E7%BB%84-%E5%90%88%E5%94%B1%E9%98%9F%E5%BD%A2"><span class="nav-text"> P1091 [NOIP 2004 提高组] 合唱队形</span></a></li></ol></div>
+            <div class="post-toc animated"><ol class="nav"><li class="nav-item nav-level-1"><a class="nav-link" href="#p1004-noip-2000-%E6%8F%90%E9%AB%98%E7%BB%84-%E6%96%B9%E6%A0%BC%E5%8F%96%E6%95%B0"><span class="nav-text"> P1004 [NOIP 2000 提高组] 方格取数</span></a></li><li class="nav-item nav-level-1"><a class="nav-link" href="#b3637-%E6%9C%80%E9%95%BF%E4%B8%8A%E5%8D%87%E5%AD%90%E5%BA%8F%E5%88%97"><span class="nav-text"> B3637 最长上升子序列</span></a></li><li class="nav-item nav-level-1"><a class="nav-link" href="#p2782-%E5%8F%8B%E5%A5%BD%E5%9F%8E%E5%B8%82"><span class="nav-text"> P2782 友好城市</span></a><ol class="nav-child"><li class="nav-item nav-level-2"><a class="nav-link" href="#%E8%B4%AA%E5%BF%83%E6%B3%95%E6%B1%82%E8%A7%A3lis"><span class="nav-text"> 贪心法求解LIS</span></a><ol class="nav-child"><li class="nav-item nav-level-3"><a class="nav-link" href="#%E4%BC%98%E5%8C%96dp%E6%80%9D%E8%B7%AF%E4%BA%A4%E6%8D%A2%E7%8A%B6%E6%80%81%E4%B8%8E%E7%8A%B6%E6%80%81%E5%80%BC"><span class="nav-text"> 优化DP思路：交换状态与状态值</span></a></li></ol></li></ol></li><li class="nav-item nav-level-1"><a class="nav-link" href="#p1091-noip-2004-%E6%8F%90%E9%AB%98%E7%BB%84-%E5%90%88%E5%94%B1%E9%98%9F%E5%BD%A2"><span class="nav-text"> P1091 [NOIP 2004 提高组] 合唱队形</span></a></li><li class="nav-item nav-level-1"><a class="nav-link" href="#p1020-noip-1999-%E6%8F%90%E9%AB%98%E7%BB%84-%E5%AF%BC%E5%BC%B9%E6%8B%A6%E6%88%AA"><span class="nav-text"> P1020 [NOIP 1999 提高组] 导弹拦截</span></a></li></ol></div>
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@@ -223,7 +223,7 @@ <h1 class="post-title" itemprop="name headline">
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       <span class="post-meta-item-text">更新于</span>
-      <time title="修改时间：2025-03-08 09:17:51" itemprop="dateModified" datetime="2025-03-08T09:17:51+08:00">2025-03-08</time>
+      <time title="修改时间：2025-03-09 14:20:42" itemprop="dateModified" datetime="2025-03-09T14:20:42+08:00">2025-03-09</time>
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@@ -271,6 +271,15 @@ <h1 id="p1091-noip-2004-提高组-合唱队形"><a class="markdownIt-Anchor" hre
 合唱队形可以看成求两边LIS之和的最大值。此时总人数减去LIS之和的最大值，就是最少出列队员数。<br />
 这里要注意当前index+1的值才是正确的长度。len标记的是数组的总长度，但是index会动态更新寻找更小值并做替换。当index找到最小值时，后面的更大值是在index以前的，不属于当前下标i+1结尾的IS长度。</p>
 <figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="type">int</span>[] gdUp = <span class="keyword">new</span> <span class="title class_">int</span>[N];</span><br><span class="line"><span class="type">int</span> <span class="variable">lenUp</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"><span class="type">int</span>[] lenUps = <span class="keyword">new</span> <span class="title class_">int</span>[N];</span><br><span class="line"><span class="keyword">for</span> (<span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> <span class="number">0</span>; i &lt; N; i += <span class="number">1</span>) &#123;</span><br><span class="line">  <span class="type">int</span> <span class="variable">index</span> <span class="operator">=</span> lowerBound(gdUp, lenUp-<span class="number">1</span>, members[i]);</span><br><span class="line">  <span class="keyword">if</span> (index == lenUp) &#123;</span><br><span class="line">    gdUp[lenUp++] = members[i];</span><br><span class="line">  &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">    gdUp[index] = members[i];</span><br><span class="line">  &#125;</span><br><span class="line">  lenUps[i] = index + <span class="number">1</span>;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="type">int</span>[] gdDown = <span class="keyword">new</span> <span class="title class_">int</span>[N];</span><br><span class="line"><span class="type">int</span> <span class="variable">lenDown</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"><span class="type">int</span>[] lenDowns = <span class="keyword">new</span> <span class="title class_">int</span>[N];</span><br><span class="line"><span class="keyword">for</span> (<span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> N-<span class="number">1</span>; i &gt;= <span class="number">0</span>; i -= <span class="number">1</span>) &#123;</span><br><span class="line">  <span class="type">int</span> <span class="variable">index</span> <span class="operator">=</span> lowerBound(gdDown, lenDown-<span class="number">1</span>, members[i]);</span><br><span class="line">  <span class="keyword">if</span> (index == lenDown) &#123;</span><br><span class="line">    gdDown[lenDown++] = members[i];</span><br><span class="line">  &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">    gdDown[index] = members[i];</span><br><span class="line">  &#125;</span><br><span class="line">  lenDowns[i] = index + <span class="number">1</span>;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="type">int</span> <span class="variable">max</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line"><span class="keyword">for</span> (<span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> <span class="number">0</span>; i &lt; N; i += <span class="number">1</span>) &#123;</span><br><span class="line">  <span class="keyword">if</span> (lenUps[i] + lenDowns[i] - <span class="number">1</span> &gt; max) &#123;</span><br><span class="line">    max = lenUps[i] + lenDowns[i] - <span class="number">1</span>;</span><br><span class="line">  &#125;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">System.out.println(N - max);</span><br></pre></td></tr></table></figure>
+<h1 id="p1020-noip-1999-提高组-导弹拦截"><a class="markdownIt-Anchor" href="#p1020-noip-1999-提高组-导弹拦截"></a> P1020 [NOIP 1999 提高组] 导弹拦截</h1>
+<p>#最长不递增子序列<br />
+做这题各种WA让我非常confusing，仔细研究后发现是我没有理解导弹拦截的规则（可以拦截相等高度！），真所谓“失之毫厘，谬以千里”。原理其实很简单：</p>
+<ol>
+<li>导弹系统可以拦截的最多导弹数，是一个<strong>最长不严格递减子序列</strong>（导弹高度不需要严格递减、可以相等），在题目要求的数据规模下，必须使用贪心+二分解法，转化为逆序求<strong>最长不严格递增子序列</strong>。</li>
+<li>最少需要多少导弹拦截系统？一个系统只能拦截比前一个导弹高度更低的导弹，那么每出现一个比之前所有高度都更高的导弹，之前的系统都不能拦截。这就是LIS！</li>
+</ol>
+<figure class="highlight java"><table><tr><td class="code"><pre><span class="line">&#123;</span><br><span class="line">    ... <span class="comment">// 处理输入</span></span><br><span class="line">    </span><br><span class="line">    <span class="type">int</span>[] gdDown = <span class="keyword">new</span> <span class="title class_">int</span>[N];</span><br><span class="line">    <span class="type">int</span> <span class="variable">maxMissile</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">for</span> (<span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> N-<span class="number">1</span>; i &gt;= <span class="number">0</span>; i -= <span class="number">1</span>) &#123;</span><br><span class="line">        <span class="type">int</span> <span class="variable">index</span> <span class="operator">=</span> upperBound(gdDown, maxMissile - <span class="number">1</span>, nums[i]);</span><br><span class="line">        <span class="keyword">if</span> (index == maxMissile) &#123;</span><br><span class="line">            gdDown[maxMissile++] = nums[i];</span><br><span class="line">        &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">            gdDown[index] = nums[i];</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="type">int</span>[] gdUp = <span class="keyword">new</span> <span class="title class_">int</span>[N];</span><br><span class="line">    <span class="type">int</span> <span class="variable">numSystems</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">for</span> (<span class="type">int</span> <span class="variable">i</span> <span class="operator">=</span> <span class="number">0</span>; i &lt; N; i += <span class="number">1</span>) &#123;</span><br><span class="line">        <span class="type">int</span> <span class="variable">index</span> <span class="operator">=</span> lowerBound(gdUp, numSystems - <span class="number">1</span>, nums[i]);</span><br><span class="line">        <span class="keyword">if</span> (index == numSystems) &#123;</span><br><span class="line">            gdUp[numSystems++] = nums[i];</span><br><span class="line">        &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">            gdUp[index] = nums[i];</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    System.out.println(maxMissile + <span class="string">&quot;\n&quot;</span> + numSystems);       </span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="comment">// 找到第一个大于该数（不管有没有找到，允许gd里的数重复）的位置</span></span><br><span class="line"><span class="keyword">private</span> <span class="keyword">static</span> <span class="type">int</span> <span class="title function_">upperBound</span><span class="params">(<span class="type">int</span>[] nums, <span class="type">int</span> end, <span class="type">int</span> target)</span> &#123;</span><br><span class="line">    <span class="type">int</span> <span class="variable">start</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">while</span> (start &lt;= end) &#123;</span><br><span class="line">        <span class="type">int</span> <span class="variable">mid</span> <span class="operator">=</span> start + (end - start) / <span class="number">2</span>;</span><br><span class="line">        <span class="keyword">if</span> (nums[mid] &lt;= target) &#123; </span><br><span class="line">            start = mid + <span class="number">1</span>;</span><br><span class="line">        &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">            end = mid - <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> start;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="comment">// 找到第一个大于（没找到，在这个位置插入）/等于（找到，在这个位置替换）该数的位置</span></span><br><span class="line"><span class="keyword">private</span> <span class="keyword">static</span> <span class="type">int</span> <span class="title function_">lowerBound</span><span class="params">(<span class="type">int</span>[] nums, <span class="type">int</span> end, <span class="type">int</span> target)</span> &#123;</span><br><span class="line">    <span class="type">int</span> <span class="variable">start</span> <span class="operator">=</span> <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">while</span> (start &lt;= end) &#123;</span><br><span class="line">        <span class="type">int</span> <span class="variable">mid</span> <span class="operator">=</span> start + (end - start) / <span class="number">2</span>;</span><br><span class="line">        <span class="keyword">if</span> (nums[mid] &gt;= target) &#123;</span><br><span class="line">            end = mid - <span class="number">1</span>;</span><br><span class="line">        &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">            start = mid + <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> start;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
+
     </div>