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chapter_dynamic_programming/dp_problem_features/index.html

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+      <a href="../unbounded_knapsack_problem/" class="md-nav__link">
+        13.5. &nbsp; 完全背包问题（New）
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 <p>给定一个楼梯，你每步可以上 <span class="arithmatex">\(1\)</span> 阶或者 <span class="arithmatex">\(2\)</span> 阶，每一阶楼梯上都贴有一个非负整数，表示你在该台阶所需要付出的代价。给定一个非负整数数组 <span class="arithmatex">\(cost\)</span> ，其中 <span class="arithmatex">\(cost[i]\)</span> 表示在第 <span class="arithmatex">\(i\)</span> 个台阶需要付出的代价，<span class="arithmatex">\(cost[0]\)</span> 为地面起始点。请计算最少需要付出多少代价才能到达顶部？</p>
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 <p>如下图所示，若第 <span class="arithmatex">\(1\)</span> , <span class="arithmatex">\(2\)</span> , <span class="arithmatex">\(3\)</span> 阶的代价分别为 <span class="arithmatex">\(1\)</span> , <span class="arithmatex">\(10\)</span> , <span class="arithmatex">\(1\)</span> ，则从地面爬到第 <span class="arithmatex">\(3\)</span> 阶的最小代价为 <span class="arithmatex">\(2\)</span> 。</p>
-<p><img alt="爬到第 3 阶的最小代价" src="../intro_to_dynamic_programming.assets/min_cost_cs_example.png" /></p>
+<p><img alt="爬到第 3 阶的最小代价" src="../dp_problem_features.assets/min_cost_cs_example.png" /></p>
 <p align="center"> Fig. 爬到第 3 阶的最小代价 </p>
 
 <p>设 <span class="arithmatex">\(dp[i]\)</span> 为爬到第 <span class="arithmatex">\(i\)</span> 阶累计付出的代价，由于第 <span class="arithmatex">\(i\)</span> 阶只可能从 <span class="arithmatex">\(i - 1\)</span> 阶或 <span class="arithmatex">\(i - 2\)</span> 阶走来，因此 <span class="arithmatex">\(dp[i]\)</span> 只可能等于 <span class="arithmatex">\(dp[i - 1] + cost[i]\)</span> 或 <span class="arithmatex">\(dp[i - 2] + cost[i]\)</span> 。为了尽可能减少代价，我们应该选择两者中较小的那一个，即：</p>
@@ -2341,7 +2357,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
 </div>
 </div>
 </div>
-<p><img alt="爬楼梯最小代价的动态规划过程" src="../intro_to_dynamic_programming.assets/min_cost_cs_dp.png" /></p>
+<p><img alt="爬楼梯最小代价的动态规划过程" src="../dp_problem_features.assets/min_cost_cs_dp.png" /></p>
 <p align="center"> Fig. 爬楼梯最小代价的动态规划过程 </p>
 
 <p>这道题同样也可以进行状态压缩，将一维压缩至零维，使得空间复杂度从 <span class="arithmatex">\(O(n)\)</span> 降低至 <span class="arithmatex">\(O(1)\)</span> 。</p>
@@ -2446,7 +2462,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
 <p>给定一个共有 <span class="arithmatex">\(n\)</span> 阶的楼梯，你每步可以上 <span class="arithmatex">\(1\)</span> 阶或者 <span class="arithmatex">\(2\)</span> 阶，<strong>但不能连续两轮跳 <span class="arithmatex">\(1\)</span> 阶</strong>，请问有多少种方案可以爬到楼顶。</p>
 </div>
 <p>例如，爬上第 <span class="arithmatex">\(3\)</span> 阶仅剩 <span class="arithmatex">\(2\)</span> 种可行方案，其中连续三次跳 <span class="arithmatex">\(1\)</span> 阶的方案不满足约束条件，因此被舍弃。</p>
-<p><img alt="带约束爬到第 3 阶的方案数量" src="../intro_to_dynamic_programming.assets/climbing_stairs_constraint_example.png" /></p>
+<p><img alt="带约束爬到第 3 阶的方案数量" src="../dp_problem_features.assets/climbing_stairs_constraint_example.png" /></p>
 <p align="center"> Fig. 带约束爬到第 3 阶的方案数量 </p>
 
 <p>在该问题中，<strong>下一步选择不能由当前状态（当前楼梯阶数）独立决定，还和前一个状态（上轮楼梯阶数）有关</strong>。如果上一轮是跳 <span class="arithmatex">\(1\)</span> 阶上来的，那么下一轮就必须跳 <span class="arithmatex">\(2\)</span> 阶。</p>
@@ -2463,7 +2479,7 @@ dp[i, 1] = dp[i-1, 2] \\
 dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
 \end{cases}
 \]</div>
-<p><img alt="考虑约束下的递推关系" src="../intro_to_dynamic_programming.assets/climbing_stairs_constraint_state_transfer.png" /></p>
+<p><img alt="考虑约束下的递推关系" src="../dp_problem_features.assets/climbing_stairs_constraint_state_transfer.png" /></p>
 <p align="center"> Fig. 考虑约束下的递推关系 </p>
 
 <p>最终，返回 <span class="arithmatex">\(dp[n, 1] + dp[n, 2]\)</span> 即可，两者之和代表爬到第 <span class="arithmatex">\(n\)</span> 阶的方案总数。</p>