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<li class="md-nav__item">
<a href="../unbounded_knapsack_problem/" class="md-nav__link">
13.5. &nbsp; 完全背包问题（New）
</a>
</li>
</ul>
</nav>
</li>
@ -2325,12 +2341,12 @@ dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
<p class="admonition-title">Note</p>
<p>边界条件即初始状态，在搜索中用于剪枝，在动态规划中用于初始化 <span class="arithmatex">\(dp\)</span> 表。状态转移顺序的核心是要保证在计算当前问题时，所有它依赖的更小子问题都已经被正确地计算出来。</p>
</div>
<p>最后，我们基于以上结果实现解法即可。熟练度较高同学可以直接写出动态规划解法，初学者可以按照“暴力搜索 <span class="arithmatex">\(\rightarrow\)</span> 记忆化搜索 <span class="arithmatex">\(\rightarrow\)</span> 动态规划” 的顺序实现。</p>
<p>接下来，我们就可以实现动态规划代码了。然而，由于子问题分解是一种从顶至底的思想，因此按照“暴力搜索 <span class="arithmatex">\(\rightarrow\)</span> 记忆化搜索 <span class="arithmatex">\(\rightarrow\)</span> 动态规划”的顺序实现更加符合思维习惯。</p>
<h2 id="1333">13.3.3. &nbsp; 方法一：暴力搜索<a class="headerlink" href="#1333" title="Permanent link">&para;</a></h2>
<p>从状态 <span class="arithmatex">\([i, j]\)</span> 开始搜索，不断分解为更小的状态 <span class="arithmatex">\([i-1, j]\)</span> 和 <span class="arithmatex">\([i, j-1]\)</span> ，包括以下递归要素：</p>
<ul>
<li><strong>递归参数</strong>：状态 <span class="arithmatex">\([i, j]\)</span> ；<strong>返回值</strong>：从 <span class="arithmatex">\([0, 0]\)</span> 到 <span class="arithmatex">\([i, j]\)</span> 的最小路径和 <span class="arithmatex">\(dp[i, j]\)</span> ；</li>
<li><strong>终止条件</strong>：当 <span class="arithmatex">\(i = 0\)</span> 且 <span class="arithmatex">\(j = 0\)</span> 时，返回代价 <span class="arithmatex">\(grid[0][0]\)</span> ；</li>
<li><strong>终止条件</strong>：当 <span class="arithmatex">\(i = 0\)</span> 且 <span class="arithmatex">\(j = 0\)</span> 时，返回代价 <span class="arithmatex">\(grid[0, 0]\)</span> ；</li>
<li><strong>剪枝</strong>：当 <span class="arithmatex">\(i &lt; 0\)</span> 时或 <span class="arithmatex">\(j &lt; 0\)</span> 时索引越界，此时返回代价 <span class="arithmatex">\(+\infty\)</span> ，代表不可行；</li>
</ul>
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