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button">RSS</a></div></div></header><cover></cover><main><div class="post-content"><div class="post-title"><h1 class="post-title__text">2021 CSP-S 复赛游记</h1><div class="post-title__meta"><a href="/archives/2021/10/" class="post-meta__date button">2021-10-24</a> <span class="separate-dot"></span> <a href="/categories/%E6%B0%B4/" class="button"><span class="post-meta__cats">水</span></a><style>.post-meta__pv{color:var(--t-l);visibility:hidden;opacity:0;transition:.2s}</style><span class="separate-dot"></span> <span class="post-meta__pv"></span></div></div><aside class="post-side"><div class="post-side__toc"><div class="toc-title">文章目录</div><ol class="toc"><li class="toc-item toc-level-2"><a class="toc-link" href="#day-0"><span class="toc-text">Day 0</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#day-1"><span class="toc-text">Day 1</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#t1-%E5%BB%8A%E6%A1%A5%E5%88%86%E9%85%8D-airport"><span class="toc-text">T1 廊桥分配 airport</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#t3-%E5%9B%9E%E6%96%87-palin"><span class="toc-text">T3 回文 palin</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#t4-%E4%BA%A4%E9%80%9A%E8%A7%84%E5%88%92-traffic"><span class="toc-text">T4 交通规划 traffic</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#t2-%E6%8B%AC%E5%8F%B7%E5%BA%8F%E5%88%97-bracket"><span class="toc-text">T2 括号序列 bracket</span></a></li></ol></li><li class="toc-item toc-level-2"><a class="toc-link" href="#20211024"><span class="toc-text">2021.10.24</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%80%BB%E7%BB%93"><span class="toc-text">总结</span></a></li></ol></div></aside><a class="btn-toc button" id="btn-toc" tabindex="0"><svg viewBox="0 0 1024 1024" width="32" height="32" xmlns="http://www.w3.org/2000/svg"><path d="M128 256h64V192H128zM320 256h576V192H320zM128 544h64v-64H128zM320 544h576v-64H320zM128 832h64v-64H128zM320 832h576v-64H320z" fill="currentColor"></path></svg></a><div class="toc-menus" id="toc-menus"><div class="toc-title">文章目录</div><ol class="toc"><li class="toc-item toc-level-2"><a class="toc-link" href="#day-0"><span class="toc-text">Day 0</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#day-1"><span class="toc-text">Day 1</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#t1-%E5%BB%8A%E6%A1%A5%E5%88%86%E9%85%8D-airport"><span class="toc-text">T1 廊桥分配 airport</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#t3-%E5%9B%9E%E6%96%87-palin"><span class="toc-text">T3 回文 palin</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#t4-%E4%BA%A4%E9%80%9A%E8%A7%84%E5%88%92-traffic"><span class="toc-text">T4 交通规划 traffic</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#t2-%E6%8B%AC%E5%8F%B7%E5%BA%8F%E5%88%97-bracket"><span class="toc-text">T2 括号序列 bracket</span></a></li></ol></li><li class="toc-item toc-level-2"><a class="toc-link" href="#20211024"><span class="toc-text">2021.10.24</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%80%BB%E7%BB%93"><span class="toc-text">总结</span></a></li></ol></div><article class="post post__with-toc card"><div class="post__header"><img alt="Cover Image" class="lazy" src="https://hans362-img.oss.0vv0.top/2019/11/03/noi.png?width=1920" srcset="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAABlBMVEXMzMyWlpYU2uzLAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAACklEQVQImWNgAAAAAgAB9HFkpgAAAABJRU5ErkJggg==" data-srcset="https://hans362-img.oss.0vv0.top/2019/11/03/noi.png?width=1920"><div class="post__expire" id="post-expired-notify"><p><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" viewBox="0 0 16 16" style="fill:#f5a623;stroke:#f5a623"><path fill-rule="evenodd" d="M8.893 1.5c-.183-.31-.52-.5-.887-.5s-.703.19-.886.5L.138 13.499a.98.98 0 0 0 0 1.001c.193.31.53.501.886.501h13.964c.367 0 .704-.19.877-.5a1.03 1.03 0 0 0 .01-1.002L8.893 1.5zm.133 11.497H6.987v-2.003h2.039v2.003zm0-3.004H6.987V5.987h2.039v4.006z"></path></svg> 本文最后更新于 <span id="expire-date"></span> 天前，文中部分描述可能已经过时。</p></div><script>(()=>{var e=Date.parse("2021-10-24"),t=(new Date).getTime(),t=Math.floor((t-e)/864e5);120<=t&&(document.querySelectorAll("#expire-date")[0].innerHTML=t,document.querySelectorAll("#post-expired-notify")[0].style.display="block")})()</script></div><div class="post__content"><html><head><script>var meting_api="https://api-v2.hans362.cn/vip/?server=:server&type=:type&id=:id&r=:r"</script><script class="meting-secondary-script-marker" src="/js/Meting.min.js"></script></head><body><p>昨天，也就是2021年10月23日，我刚刚考完了本年度的 CSP-S 第二轮。晚上六点多走出考场时，我人已经麻了。欲知发生了甚么事，请继续阅读本文。</p><p>P.S. 去年复赛游记请移步：<a href="https://blog.hans362.cn/post/2020-csp-s-2/">2020 CSP-S 复赛游记</a></p><span id="more"></span><h2 id="day-0"><a class="markdownIt-Anchor" href="#day-0"></a> Day 0</h2><p>复赛那天恰逢成人高考，原以为我校应该做不了 CSP 认证点，和初赛一样要跑很远去华二考试，结果一看准考证，认证点竟然就在自己学校？！瞬间狂喜，而且机房也在以前刷题的机房，监考老师就是自己教练。</p><p>考试前一天下午因为成人高考布置考场三点多就放学了，回到家大致梳理了一下常用模版的实现思路。由于今年9月1日起 NOI 系列活动启用全新的技术标准，采用 C++14 标准，放开了大部分原本禁用的下划线开头的函数，<code>pbds</code>也可以使用，因此又现学了一下一些原本禁用的函数和库。</p><h2 id="day-1"><a class="markdownIt-Anchor" href="#day-1"></a> Day 1</h2><p>下午提前45分钟到达学校，看到校门口好多人于是想也没想就跟在后面排队测温，快排到我时我瞥了一眼工作人员的胸牌，一眼就看到大大的 SHMEEA 的标志。嗯？怎么是上海市教育考试院的牌子？再一看别人手里拿的准考证，然后瞬间明白了我混进了成人高考的队伍。当然也无伤大雅，进去后直奔科技楼，终于看到教练在门口负责收集考试安全承诺书。因为时间还早就先去底楼的实验室开了台电脑练习一下手感，提前十五分钟再进楼上的考场。</p><p>到点了公布解压密码，一次输对，解压、建文件夹一气呵成，毕竟都打了两年比赛了。然后开始看题，发现今年可以吸氧（开 O2 优化）诶，STL 可以放心用了。吸取去年 T1 的教训，我把4道题都看完了。然而看完我就慌了，因为我根本找不到签到题，就只好先从 T1 开始。</p><h3 id="t1-廊桥分配-airport"><a class="markdownIt-Anchor" href="#t1-廊桥分配-airport"></a> T1 廊桥分配 airport</h3><p>尝试抽象了一下数学模型，感觉和线段覆盖有关？莫非是线段树？那我可不会打。思索了半小时没啥正经的思路，暴力做法倒是想到了，也就是将每架飞机的起飞降落都分别看成一个事件，用结构体维护事件，上来先把事件按时序用 <code>sort()</code> 排列，然后对于每一种分配方案再模拟时间轴，统计停靠廊桥的飞机数量，最后取其中的最大值。当然我也很清楚这样的做法时间复杂度是 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mi>l</mi><mi>o</mi><mi>g</mi><mi>n</mi><mo stretchy="false">(</mo><msub><mi>m</mi><mn>1</mn></msub><mo>+</mo><msub><mi>m</mi><mn>2</mn></msub><mo stretchy="false">)</mo><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">O(nlogn(m_1+m_2))</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-.25em"></span><span class="mord mathnormal" style="margin-right:.02778em">O</span><span class="mopen">(</span><span class="mord mathnormal">n</span><span class="mord mathnormal" style="margin-right:.01968em">l</span><span class="mord mathnormal">o</span><span class="mord mathnormal" style="margin-right:.03588em">g</span><span class="mord mathnormal">n</span><span class="mopen">(</span><span class="mord"><span class="mord mathnormal">m</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:.30110799999999993em"><span style="top:-2.5500000000000003em;margin-left:0;margin-right:.05em"><span class="pstrut" style="height:2.7em"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">1</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:.15em"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:.2222222222222222em"></span><span class="mbin">+</span><span class="mspace" style="margin-right:.2222222222222222em"></span></span><span class="base"><span class="strut" style="height:1em;vertical-align:-.25em"></span><span class="mord"><span class="mord mathnormal">m</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:.30110799999999993em"><span style="top:-2.5500000000000003em;margin-left:0;margin-right:.05em"><span class="pstrut" style="height:2.7em"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:.15em"><span></span></span></span></span></span></span><span class="mclose">)</span><span class="mclose">)</span></span></span></span>，看了下数据范围 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:.43056em;vertical-align:0"></span><span class="mord mathnormal">n</span></span></span></span> 和 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>m</mi><mn>1</mn></msub><mo>+</mo><msub><mi>m</mi><mn>2</mn></msub></mrow><annotation encoding="application/x-tex">m_1+m_2</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:.73333em;vertical-align:-.15em"></span><span class="mord"><span class="mord mathnormal">m</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:.30110799999999993em"><span style="top:-2.5500000000000003em;margin-left:0;margin-right:.05em"><span class="pstrut" style="height:2.7em"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">1</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:.15em"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:.2222222222222222em"></span><span class="mbin">+</span><span class="mspace" style="margin-right:.2222222222222222em"></span></span><span class="base"><span class="strut" style="height:.58056em;vertical-align:-.15em"></span><span class="mord"><span class="mord mathnormal">m</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:.30110799999999993em"><span style="top:-2.5500000000000003em;margin-left:0;margin-right:.05em"><span class="pstrut" style="height:2.7em"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:.15em"><span></span></span></span></span></span></span></span></span></span> 都要到 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mi>e</mi><mn>5</mn></mrow><annotation encoding="application/x-tex">1e5</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:.64444em;vertical-align:0"></span><span class="mord">1</span><span class="mord mathnormal">e</span><span class="mord">5</span></span></span></span>，明显会爆炸。然而已经过去半小时了，实在没什么更好的思路了，只能先暴力开搞。半小时写完样例都通过了，可见样例非常水。</p><h3 id="t3-回文-palin"><a class="markdownIt-Anchor" href="#t3-回文-palin"></a> T3 回文 palin</h3><p>因为 T2 实在没啥想法甚至连暴力都不知道咋写，果断跳 T3。这题也没啥正经做法的思路，看到输出字典序最小的结果考虑可以 DFS，优先走搜索树的左子树即可，当然看了下数据范围肯定炸，因此想写个 DFS 再试试看能不能剪枝。写完 DFS 过了前两个样例，考虑剪枝。发现搜索到 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>+</mo><mn>2</mn></mrow><annotation encoding="application/x-tex">n+2</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:.66666em;vertical-align:-.08333em"></span><span class="mord mathnormal">n</span><span class="mspace" style="margin-right:.2222222222222222em"></span><span class="mbin">+</span><span class="mspace" style="margin-right:.2222222222222222em"></span></span><span class="base"><span class="strut" style="height:.64444em;vertical-align:0"></span><span class="mord">2</span></span></span></span> 层就可以开始判断是否回文了，如果不是就直接回溯，即</p><pre><code class="hljs apache"><span class="hljs-attribute">if</span> (k>n+<span class="hljs-number">1</span> && (b[k-<span class="hljs-number">1</span>]!=b[<span class="hljs-number">2</span>*n+<span class="hljs-number">2</span>-k])) return;</code></pre><p>然而样例3还是炸，遂放弃。</p><h3 id="t4-交通规划-traffic"><a class="markdownIt-Anchor" href="#t4-交通规划-traffic"></a> T4 交通规划 traffic</h3><p>题看起来都费劲，而且不像去年，今年这道没有好做的部分分。显然不可做，就直接放弃了，回头去看 T2。</p><h3 id="t2-括号序列-bracket"><a class="markdownIt-Anchor" href="#t2-括号序列-bracket"></a> T2 括号序列 bracket</h3><p>这题一上来完全没往 DP 上想，我的思路是枚举每个<code>?</code>可能的值，再判断整个字符串是否合法。但是问题就出在这个判断函数应该咋写，我联想到了利用栈进行括号匹配的操作，于是大胆猜想这题也应该是这么干的，只不过还要确认连续出栈的<code>*</code>个数小于等于 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:.69444em;vertical-align:0"></span><span class="mord mathnormal" style="margin-right:.03148em">k</span></span></span></span>，然而很不幸猜错了。当样例1死活过不了之时，我才突然发现我这个算法很容易就证伪了。然后我就开始想另一种校验字符串合法的方式，完全陷入了这种思维定势导致我浪费了大量时间。</p><p>最后只剩下半小时了我才突然意识到 DP，结果当然并没有推出转移方程。无奈放弃。试图开一下虚拟机准备跑一下 T1 T3，结果学校的破电脑根本承受不住竟然直接死机了，硬盘仿佛在呼啸，最后也没测成。</p><p>到点交卷签字走人，我觉得我凉透了，应该是没有三位数了。</p><h2 id="20211024"><a class="markdownIt-Anchor" href="#20211024"></a> 2021.10.24</h2><p>一早测了下洛谷民间数据，40+0+28+0，我人没了。后来发现 T3 还有一个枝忘剪了，加了行代码直接 28pts 变 40pts，我人麻了。</p><h2 id="总结"><a class="markdownIt-Anchor" href="#总结"></a> 总结</h2><p>今年可以说更加注重思维的考察，没有去年的大模拟、卡精度之类的妖魔鬼怪，但是对于我这种缺乏思考能力的选手而言确实是挺有难度的，至少我认为比去年难多了。而我这次 T2 被自己刚开始的思路局限住了，没有充足的时间去思考 DP 做法，也值得引以为鉴。</p><p>总之这场 CSP 打得稀烂，但愿还能有机会苟进 NOIP。</p></body></html></div><div class="license"><div class="license-title">2021 CSP-S 复赛游记</div><div class="license-link"><a href="https://blog.hans362.cn/post/2021-csp-s-2/">https://blog.hans362.cn/post/2021-csp-s-2/</a></div><div class="license-meta"><div class="license-meta-item"><div class="license-meta-title">本文作者</div><div class="license-meta-text">Hans362</div></div><div class="license-meta-item"><div class="license-meta-title">最后更新</div><div class="license-meta-text">2021-10-24</div></div><div class="license-meta-item"><div class="license-meta-title">许可协议</div><div class="license-meta-text"><a href="https://creativecommons.org/licenses/by-nc-sa/4.0/deed.zh" rel="nofollow noopener noreferrer" target="_blank">CC BY-NC-SA 4.0</a></div></div></div><div>转载或引用本文时请遵守许可协议，注明出处、不得用于商业用途！</div></div><div 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