xinyin025/hello-algo
1
0
Fork 0
mirror of https://github.com/krahets/hello-algo.git synced 2026-08-11 19:30:59 +00:00
Code Issues Packages Projects Releases Wiki Activity

deploy

Browse Source
This commit is contained in:
krahets
2023-08-21 19:32:49 +08:00
parent c0f960b443
commit c359c07fe0
67 changed files with 443 additions and 442 deletions
Download Patch File Download Diff File Expand all files Collapse all files
chapter_greedy/max_capacity_problem/index.html
+7 -6
View File
@@ -3454,7 +3454,7 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
<p><img alt="初始状态" src="../max_capacity_problem.assets/max_capacity_initial_state.png" /></p>
<p align="center"> 图：初始状态 </p>
<p>我们发现，<strong>如果此时将长板 <span class="arithmatex">\(j\)</span> 向短板 <span class="arithmatex">\(i\)</span> 靠近，则容量一定变小</strong>。这是因为在移动长板 <span class="arithmatex">\(j\)</span> 后：</p>
<p>如下图所示，<strong>若此时将长板 <span class="arithmatex">\(j\)</span> 向短板 <span class="arithmatex">\(i\)</span> 靠近，则容量一定变小</strong>。这是因为在移动长板 <span class="arithmatex">\(j\)</span> 后：</p>
<ul>
<li>宽度 <span class="arithmatex">\(j-i\)</span> 肯定变小。</li>
<li>高度由短板决定，因此高度只可能不变（ <span class="arithmatex">\(i\)</span> 仍为短板）或变小（移动后的 <span class="arithmatex">\(j\)</span> 成为短板）。</li>
@@ -3462,11 +3462,12 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
<p><img alt="向内移动长板后的状态" src="../max_capacity_problem.assets/max_capacity_moving_long_board.png" /></p>
<p align="center"> 图：向内移动长板后的状态 </p>
<p>反向思考，<strong>我们只有向内收缩短板 <span class="arithmatex">\(i\)</span> ，才有可能使容量变大</strong>。因为虽然宽度一定变小，<strong>但高度可能会变大</strong>（移动后的短板 <span class="arithmatex">\(i\)</span> 可能会变长）。</p>
<p><img alt="向内移动长板后的状态" src="../max_capacity_problem.assets/max_capacity_moving_short_board.png" /></p>
<p align="center"> 图：向内移动长板后的状态 </p>
<p>反向思考，<strong>我们只有向内收缩短板 <span class="arithmatex">\(i\)</span> ，才有可能使容量变大</strong>。因为虽然宽度一定变小，<strong>但高度可能会变大</strong>（移动后的短板 <span class="arithmatex">\(i\)</span> 可能会变长）。例如在下图中，移动短板后面积变大。</p>
<p><img alt="向内移动短板后的状态" src="../max_capacity_problem.assets/max_capacity_moving_short_board.png" /></p>
<p align="center"> 图：向内移动短板后的状态 </p>
<p>由此便可推出本题的贪心策略：</p>
<p>由此便可推出本题的贪心策略：初始化两指针分裂容器两端，每轮向内收缩短板对应的指针，直至两指针相遇。</p>
<p>下图展示了贪心策略的执行过程。</p>
<ol>
<li>初始状态下，指针 <span class="arithmatex">\(i\)</span> , <span class="arithmatex">\(j\)</span> 分列与数组两端。</li>
<li>计算当前状态的容量 <span class="arithmatex">\(cap[i, j]\)</span> ，并更新最大容量。</li>
@@ -3694,7 +3695,7 @@ cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
</div>
<h3 id="3">3. &nbsp; 正确性证明<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
<p>之所以贪心比穷举更快，是因为每轮的贪心选择都会“跳过”一些状态。</p>
<p>比如在状态 <span class="arithmatex">\(cap[i, j]\)</span> 下，<span class="arithmatex">\(i\)</span> 为短板、<span class="arithmatex">\(j\)</span> 为长板。若贪心地将短板 <span class="arithmatex">\(i\)</span> 向内移动一格，会导致以下状态被“跳过”。<strong>这意味着之后无法验证这些状态的容量大小</strong>。</p>
<p>比如在状态 <span class="arithmatex">\(cap[i, j]\)</span> 下，<span class="arithmatex">\(i\)</span> 为短板、<span class="arithmatex">\(j\)</span> 为长板。若贪心地将短板 <span class="arithmatex">\(i\)</span> 向内移动一格，会导致下图所示的状态被“跳过”。<strong>这意味着之后无法验证这些状态的容量大小</strong>。</p>
<div class="arithmatex">\[
cap[i, i+1], cap[i, i+2], \dots, cap[i, j-2], cap[i, j-1]
\]</div>