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leetcode-problemset/leetcode-cn/problem (Chinese)/覆盖网格的最少传感器数目 [minimum-sensors-to-cover-grid].html
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<p>给你一个 <code>n × m</code> 的网格和一个整数 <code>k</code></p>
<p>一个放置在单元格 <code>(r, c)</code> 的传感器可以覆盖所有与 <code>(r, c)</code>&nbsp;<strong>切比雪夫距离</strong><strong>不超过</strong> <code>k</code> 的单元格。</p>
<p>两个单元格 <code>(r<sub>1</sub>, c<sub>1</sub>)</code><code>(r<sub>2</sub>, c<sub>2</sub>)</code> 之间的&nbsp;<strong>切比雪夫距离&nbsp;</strong><code>max(|r<sub>1</sub> r<sub>2</sub>|,|c<sub>1</sub> c<sub>2</sub>|)</code></p>
<p>你的任务是返回覆盖整个网格所需传感器的&nbsp;<strong>最少&nbsp;</strong>数量。</p>
<p>&nbsp;</p>
<p><strong class="example">示例 1:</strong></p>
<div class="example-block">
<p><strong>输入:</strong> <span class="example-io">n = 5, m = 5, k = 1</span></p>
<p><strong>输出:</strong> <span class="example-io">4</span></p>
<p><strong>解释:</strong></p>
<p>在位置 <code>(0, 3)</code><code>(1, 0)</code><code>(3, 3)</code><code>(4, 1)</code> 放置传感器可以确保网格中的每个单元格都被覆盖。因此,答案是 4。</p>
</div>
<p><strong class="example">示例 2:</strong></p>
<div class="example-block">
<p><strong>输入:</strong> <span class="example-io">n = 2, m = 2, k = 2</span></p>
<p><strong>输出:</strong> <span class="example-io">1</span></p>
<p><strong>解释:</strong></p>
<p><code>k = 2</code> 时,无论传感器放在哪个位置,单个传感器都可以覆盖整个 <code>2 * 2</code> 的网格。因此,答案是 1。</p>
</div>
<p>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li><code>1 &lt;= n &lt;= 10<sup>3</sup></code></li>
<li><code>1 &lt;= m &lt;= 10<sup>3</sup></code></li>
<li><code>0 &lt;= k &lt;= 10<sup>3</sup></code></li>
</ul>