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# 14.2 &nbsp; 动态规划问题特性

在上一节中，我们学习了动态规划是如何通过子问题分解来求解原问题的。实际上，子问题分解是一种通用的算法思路，在分治、动态规划、回溯中的侧重点不同。

- 分治算法递归地将原问题划分为多个相互独立的子问题，直至最小子问题，并在回溯中合并子问题的解，最终得到原问题的解。
- 动态规划也对问题进行递归分解，但与分治算法的主要区别是，动态规划中的子问题是相互依赖的，在分解过程中会出现许多重叠子问题。
- 回溯算法在尝试和回退中穷举所有可能的解，并通过剪枝避免不必要的搜索分支。原问题的解由一系列决策步骤构成，我们可以将每个决策步骤之前的子序列看作一个子问题。

实际上，动态规划常用来求解最优化问题，它们不仅包含重叠子问题，还具有另外两大特性：最优子结构、无后效性。

## 14.2.1 &nbsp; 最优子结构

我们对爬楼梯问题稍作改动，使之更加适合展示最优子结构概念。

!!! question "爬楼梯最小代价"

    给定一个楼梯，你每步可以上 $1$ 阶或者 $2$ 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 $cost$ ，其中 $cost[i]$ 表示在第 $i$ 个台阶需要付出的代价，$cost[0]$ 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？

如图 14-6 所示，若第 $1$、$2$、$3$ 阶的代价分别为 $1$、$10$、$1$ ，则从地面爬到第 $3$ 阶的最小代价为 $2$ 。

![爬到第 3 阶的最小代价](dp_problem_features.assets/min_cost_cs_example.png){ class="animation-figure" }

<p align="center"> 图 14-6 &nbsp; 爬到第 3 阶的最小代价 </p>

设 $dp[i]$ 为爬到第 $i$ 阶累计付出的代价，由于第 $i$ 阶只可能从 $i - 1$ 阶或 $i - 2$ 阶走来，因此 $dp[i]$ 只可能等于 $dp[i - 1] + cost[i]$ 或 $dp[i - 2] + cost[i]$ 。为了尽可能减少代价，我们应该选择两者中较小的那一个：

$$
dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
$$

这便可以引出最优子结构的含义：**原问题的最优解是从子问题的最优解构建得来的**。

本题显然具有最优子结构：我们从两个子问题最优解 $dp[i-1]$ 和 $dp[i-2]$ 中挑选出较优的那一个，并用它构建出原问题 $dp[i]$ 的最优解。

那么，上一节的爬楼梯题目有没有最优子结构呢？它的目标是求解方案数量，看似是一个计数问题，但如果换一种问法：“求解最大方案数量”。我们意外地发现，**虽然题目修改前后是等价的，但最优子结构浮现出来了**：第 $n$ 阶最大方案数量等于第 $n-1$ 阶和第 $n-2$ 阶最大方案数量之和。所以说，最优子结构的解释方式比较灵活，在不同问题中会有不同的含义。

根据状态转移方程，以及初始状态 $dp[1] = cost[1]$ 和 $dp[2] = cost[2]$ ，我们就可以得到动态规划代码：

=== "Python"

    ```python title="min_cost_climbing_stairs_dp.py"
    def min_cost_climbing_stairs_dp(cost: list[int]) -> int:
        """爬楼梯最小代价：动态规划"""
        n = len(cost) - 1
        if n == 1 or n == 2:
            return cost[n]
        # 初始化 dp 表，用于存储子问题的解
        dp = [0] * (n + 1)
        # 初始状态：预设最小子问题的解
        dp[1], dp[2] = cost[1], cost[2]
        # 状态转移：从较小子问题逐步求解较大子问题
        for i in range(3, n + 1):
            dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
        return dp[n]
    ```

=== "C++"

    ```cpp title="min_cost_climbing_stairs_dp.cpp"
    /* 爬楼梯最小代价：动态规划 */
    int minCostClimbingStairsDP(vector<int> &cost) {
        int n = cost.size() - 1;
        if (n == 1 || n == 2)
            return cost[n];
        // 初始化 dp 表，用于存储子问题的解
        vector<int> dp(n + 1);
        // 初始状态：预设最小子问题的解
        dp[1] = cost[1];
        dp[2] = cost[2];
        // 状态转移：从较小子问题逐步求解较大子问题
        for (int i = 3; i <= n; i++) {
            dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];
        }
        return dp[n];
    }
    ```

=== "Java"

    ```java title="min_cost_climbing_stairs_dp.java"
    /* 爬楼梯最小代价：动态规划 */
    int minCostClimbingStairsDP(int[] cost) {
        int n = cost.length - 1;
        if (n == 1 || n == 2)
            return cost[n];
        // 初始化 dp 表，用于存储子问题的解
        int[] dp = new int[n + 1];
        // 初始状态：预设最小子问题的解
        dp[1] = cost[1];
        dp[2] = cost[2];
        // 状态转移：从较小子问题逐步求解较大子问题
        for (int i = 3; i <= n; i++) {
            dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
        }
        return dp[n];
    }
    ```

=== "C#"

    ```csharp title="min_cost_climbing_stairs_dp.cs"
    /* 爬楼梯最小代价：动态规划 */
    int MinCostClimbingStairsDP(int[] cost) {
        int n = cost.Length - 1;
        if (n == 1 || n == 2)
            return cost[n];
        // 初始化 dp 表，用于存储子问题的解
        int[] dp = new int[n + 1];
        // 初始状态：预设最小子问题的解
        dp[1] = cost[1];
        dp[2] = cost[2];
        // 状态转移：从较小子问题逐步求解较大子问题
        for (int i = 3; i <= n; i++) {
            dp[i] = Math.Min(dp[i - 1], dp[i - 2]) + cost[i];
        }
        return dp[n];
    }
    ```

=== "Go"

    ```go title="min_cost_climbing_stairs_dp.go"
    /* 爬楼梯最小代价：动态规划 */
    func minCostClimbingStairsDP(cost []int) int {
        n := len(cost) - 1
        if n == 1 || n == 2 {
            return cost[n]
        }
        min := func(a, b int) int {
            if a < b {
                return a
            }
            return b
        }
        // 初始化 dp 表，用于存储子问题的解
        dp := make([]int, n+1)
        // 初始状态：预设最小子问题的解
        dp[1] = cost[1]
        dp[2] = cost[2]
        // 状态转移：从较小子问题逐步求解较大子问题
        for i := 3; i <= n; i++ {
            dp[i] = min(dp[i-1], dp[i-2]) + cost[i]
        }
        return dp[n]
    }
    ```

=== "Swift"

    ```swift title="min_cost_climbing_stairs_dp.swift"
    /* 爬楼梯最小代价：动态规划 */
    func minCostClimbingStairsDP(cost: [Int]) -> Int {
        let n = cost.count - 1
        if n == 1 || n == 2 {
            return cost[n]
        }
        // 初始化 dp 表，用于存储子问题的解
        var dp = Array(repeating: 0, count: n + 1)
        // 初始状态：预设最小子问题的解
        dp[1] = cost[1]
        dp[2] = cost[2]
        // 状态转移：从较小子问题逐步求解较大子问题
        for i in 3 ... n {
            dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
        }
        return dp[n]
    }
    ```

=== "JS"

    ```javascript title="min_cost_climbing_stairs_dp.js"
    /* 爬楼梯最小代价：动态规划 */
    function minCostClimbingStairsDP(cost) {
        const n = cost.length - 1;
        if (n === 1 || n === 2) {
            return cost[n];
        }
        // 初始化 dp 表，用于存储子问题的解
        const dp = new Array(n + 1);
        // 初始状态：预设最小子问题的解
        dp[1] = cost[1];
        dp[2] = cost[2];
        // 状态转移：从较小子问题逐步求解较大子问题
        for (let i = 3; i <= n; i++) {
            dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
        }
        return dp[n];
    }
    ```

=== "TS"

    ```typescript title="min_cost_climbing_stairs_dp.ts"
    /* 爬楼梯最小代价：动态规划 */
    function minCostClimbingStairsDP(cost: Array<number>): number {
        const n = cost.length - 1;
        if (n === 1 || n === 2) {
            return cost[n];
        }
        // 初始化 dp 表，用于存储子问题的解
        const dp = new Array(n + 1);
        // 初始状态：预设最小子问题的解
        dp[1] = cost[1];
        dp[2] = cost[2];
        // 状态转移：从较小子问题逐步求解较大子问题
        for (let i = 3; i <= n; i++) {
            dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
        }
        return dp[n];
    }
    ```

=== "Dart"

    ```dart title="min_cost_climbing_stairs_dp.dart"
    /* 爬楼梯最小代价：动态规划 */
    int minCostClimbingStairsDP(List<int> cost) {
      int n = cost.length - 1;
      if (n == 1 || n == 2) return cost[n];
      // 初始化 dp 表，用于存储子问题的解
      List<int> dp = List.filled(n + 1, 0);
      // 初始状态：预设最小子问题的解
      dp[1] = cost[1];
      dp[2] = cost[2];
      // 状态转移：从较小子问题逐步求解较大子问题
      for (int i = 3; i <= n; i++) {
        dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];
      }
      return dp[n];
    }
    ```

=== "Rust"

    ```rust title="min_cost_climbing_stairs_dp.rs"
    /* 爬楼梯最小代价：动态规划 */
    fn min_cost_climbing_stairs_dp(cost: &[i32]) -> i32 {
        let n = cost.len() - 1;
        if n == 1 || n == 2 {
            return cost[n];
        }
        // 初始化 dp 表，用于存储子问题的解
        let mut dp = vec![-1; n + 1];
        // 初始状态：预设最小子问题的解
        dp[1] = cost[1];
        dp[2] = cost[2];
        // 状态转移：从较小子问题逐步求解较大子问题
        for i in 3..=n {
            dp[i] = cmp::min(dp[i - 1], dp[i - 2]) + cost[i];
        }
        dp[n]
    }
    ```

=== "C"

    ```c title="min_cost_climbing_stairs_dp.c"
    /* 爬楼梯最小代价：动态规划 */
    int minCostClimbingStairsDP(int cost[], int costSize) {
        int n = costSize - 1;
        if (n == 1 || n == 2)
            return cost[n];
        // 初始化 dp 表，用于存储子问题的解
        int *dp = calloc(n + 1, sizeof(int));
        // 初始状态：预设最小子问题的解
        dp[1] = cost[1];
        dp[2] = cost[2];
        // 状态转移：从较小子问题逐步求解较大子问题
        for (int i = 3; i <= n; i++) {
            dp[i] = myMin(dp[i - 1], dp[i - 2]) + cost[i];
        }
        int res = dp[n];
        // 释放内存
        free(dp);
        return res;
    }
    ```

=== "Kotlin"

    ```kotlin title="min_cost_climbing_stairs_dp.kt"
    /* 爬楼梯最小代价：动态规划 */
    fun minCostClimbingStairsDP(cost: IntArray): Int {
        val n = cost.size - 1
        if (n == 1 || n == 2) return cost[n]
        // 初始化 dp 表，用于存储子问题的解
        val dp = IntArray(n + 1)
        // 初始状态：预设最小子问题的解
        dp[1] = cost[1]
        dp[2] = cost[2]
        // 状态转移：从较小子问题逐步求解较大子问题
        for (i in 3..n) {
            dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
        }
        return dp[n]
    }
    ```

=== "Ruby"

    ```ruby title="min_cost_climbing_stairs_dp.rb"
    [class]{}-[func]{min_cost_climbing_stairs_dp}
    ```

=== "Zig"

    ```zig title="min_cost_climbing_stairs_dp.zig"
    // 爬楼梯最小代价：动态规划
    fn minCostClimbingStairsDP(comptime cost: []i32) i32 {
        comptime var n = cost.len - 1;
        if (n == 1 or n == 2) {
            return cost[n];
        }
        // 初始化 dp 表，用于存储子问题的解
        var dp = [_]i32{-1} ** (n + 1);
        // 初始状态：预设最小子问题的解
        dp[1] = cost[1];
        dp[2] = cost[2];
        // 状态转移：从较小子问题逐步求解较大子问题
        for (3..n + 1) |i| {
            dp[i] = @min(dp[i - 1], dp[i - 2]) + cost[i];
        }
        return dp[n];
    }
    ```

??? pythontutor "可视化运行"

    <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20min%28dp%5Bi%20-%201%5D,%20dp%5Bi%20-%202%5D%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
    <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20min%28dp%5Bi%20-%201%5D,%20dp%5Bi%20-%202%5D%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a></div>

图 14-7 展示了以上代码的动态规划过程。

![爬楼梯最小代价的动态规划过程](dp_problem_features.assets/min_cost_cs_dp.png){ class="animation-figure" }

<p align="center"> 图 14-7 &nbsp; 爬楼梯最小代价的动态规划过程 </p>

本题也可以进行空间优化，将一维压缩至零维，使得空间复杂度从 $O(n)$ 降至 $O(1)$ ：

=== "Python"

    ```python title="min_cost_climbing_stairs_dp.py"
    def min_cost_climbing_stairs_dp_comp(cost: list[int]) -> int:
        """爬楼梯最小代价：空间优化后的动态规划"""
        n = len(cost) - 1
        if n == 1 or n == 2:
            return cost[n]
        a, b = cost[1], cost[2]
        for i in range(3, n + 1):
            a, b = b, min(a, b) + cost[i]
        return b
    ```

=== "C++"

    ```cpp title="min_cost_climbing_stairs_dp.cpp"
    /* 爬楼梯最小代价：空间优化后的动态规划 */
    int minCostClimbingStairsDPComp(vector<int> &cost) {
        int n = cost.size() - 1;
        if (n == 1 || n == 2)
            return cost[n];
        int a = cost[1], b = cost[2];
        for (int i = 3; i <= n; i++) {
            int tmp = b;
            b = min(a, tmp) + cost[i];
            a = tmp;
        }
        return b;
    }
    ```

=== "Java"

    ```java title="min_cost_climbing_stairs_dp.java"
    /* 爬楼梯最小代价：空间优化后的动态规划 */
    int minCostClimbingStairsDPComp(int[] cost) {
        int n = cost.length - 1;
        if (n == 1 || n == 2)
            return cost[n];
        int a = cost[1], b = cost[2];
        for (int i = 3; i <= n; i++) {
            int tmp = b;
            b = Math.min(a, tmp) + cost[i];
            a = tmp;
        }
        return b;
    }
    ```

=== "C#"

    ```csharp title="min_cost_climbing_stairs_dp.cs"
    /* 爬楼梯最小代价：空间优化后的动态规划 */
    int MinCostClimbingStairsDPComp(int[] cost) {
        int n = cost.Length - 1;
        if (n == 1 || n == 2)
            return cost[n];
        int a = cost[1], b = cost[2];
        for (int i = 3; i <= n; i++) {
            int tmp = b;
            b = Math.Min(a, tmp) + cost[i];
            a = tmp;
        }
        return b;
    }
    ```

=== "Go"

    ```go title="min_cost_climbing_stairs_dp.go"
    /* 爬楼梯最小代价：空间优化后的动态规划 */
    func minCostClimbingStairsDPComp(cost []int) int {
        n := len(cost) - 1
        if n == 1 || n == 2 {
            return cost[n]
        }
        min := func(a, b int) int {
            if a < b {
                return a
            }
            return b
        }
        // 初始状态：预设最小子问题的解
        a, b := cost[1], cost[2]
        // 状态转移：从较小子问题逐步求解较大子问题
        for i := 3; i <= n; i++ {
            tmp := b
            b = min(a, tmp) + cost[i]
            a = tmp
        }
        return b
    }
    ```

=== "Swift"

    ```swift title="min_cost_climbing_stairs_dp.swift"
    /* 爬楼梯最小代价：空间优化后的动态规划 */
    func minCostClimbingStairsDPComp(cost: [Int]) -> Int {
        let n = cost.count - 1
        if n == 1 || n == 2 {
            return cost[n]
        }
        var (a, b) = (cost[1], cost[2])
        for i in 3 ... n {
            (a, b) = (b, min(a, b) + cost[i])
        }
        return b
    }
    ```

=== "JS"

    ```javascript title="min_cost_climbing_stairs_dp.js"
    /* 爬楼梯最小代价：状态压缩后的动态规划 */
    function minCostClimbingStairsDPComp(cost) {
        const n = cost.length - 1;
        if (n === 1 || n === 2) {
            return cost[n];
        }
        let a = cost[1],
            b = cost[2];
        for (let i = 3; i <= n; i++) {
            const tmp = b;
            b = Math.min(a, tmp) + cost[i];
            a = tmp;
        }
        return b;
    }
    ```

=== "TS"

    ```typescript title="min_cost_climbing_stairs_dp.ts"
    /* 爬楼梯最小代价：状态压缩后的动态规划 */
    function minCostClimbingStairsDPComp(cost: Array<number>): number {
        const n = cost.length - 1;
        if (n === 1 || n === 2) {
            return cost[n];
        }
        let a = cost[1],
            b = cost[2];
        for (let i = 3; i <= n; i++) {
            const tmp = b;
            b = Math.min(a, tmp) + cost[i];
            a = tmp;
        }
        return b;
    }
    ```

=== "Dart"

    ```dart title="min_cost_climbing_stairs_dp.dart"
    /* 爬楼梯最小代价：空间优化后的动态规划 */
    int minCostClimbingStairsDPComp(List<int> cost) {
      int n = cost.length - 1;
      if (n == 1 || n == 2) return cost[n];
      int a = cost[1], b = cost[2];
      for (int i = 3; i <= n; i++) {
        int tmp = b;
        b = min(a, tmp) + cost[i];
        a = tmp;
      }
      return b;
    }
    ```

=== "Rust"

    ```rust title="min_cost_climbing_stairs_dp.rs"
    /* 爬楼梯最小代价：空间优化后的动态规划 */
    fn min_cost_climbing_stairs_dp_comp(cost: &[i32]) -> i32 {
        let n = cost.len() - 1;
        if n == 1 || n == 2 {
            return cost[n];
        };
        let (mut a, mut b) = (cost[1], cost[2]);
        for i in 3..=n {
            let tmp = b;
            b = cmp::min(a, tmp) + cost[i];
            a = tmp;
        }
        b
    }
    ```

=== "C"

    ```c title="min_cost_climbing_stairs_dp.c"
    /* 爬楼梯最小代价：空间优化后的动态规划 */
    int minCostClimbingStairsDPComp(int cost[], int costSize) {
        int n = costSize - 1;
        if (n == 1 || n == 2)
            return cost[n];
        int a = cost[1], b = cost[2];
        for (int i = 3; i <= n; i++) {
            int tmp = b;
            b = myMin(a, tmp) + cost[i];
            a = tmp;
        }
        return b;
    }
    ```

=== "Kotlin"

    ```kotlin title="min_cost_climbing_stairs_dp.kt"
    /* 爬楼梯最小代价：空间优化后的动态规划 */
    fun minCostClimbingStairsDPComp(cost: IntArray): Int {
        val n = cost.size - 1
        if (n == 1 || n == 2) return cost[n]
        var a = cost[1]
        var b = cost[2]
        for (i in 3..n) {
            val tmp = b
            b = min(a, tmp) + cost[i]
            a = tmp
        }
        return b
    }
    ```

=== "Ruby"

    ```ruby title="min_cost_climbing_stairs_dp.rb"
    [class]{}-[func]{min_cost_climbing_stairs_dp_comp}
    ```

=== "Zig"

    ```zig title="min_cost_climbing_stairs_dp.zig"
    // 爬楼梯最小代价：空间优化后的动态规划
    fn minCostClimbingStairsDPComp(cost: []i32) i32 {
        var n = cost.len - 1;
        if (n == 1 or n == 2) {
            return cost[n];
        }
        var a = cost[1];
        var b = cost[2];
        // 状态转移：从较小子问题逐步求解较大子问题
        for (3..n + 1) |i| {
            var tmp = b;
            b = @min(a, tmp) + cost[i];
            a = tmp;
        }
        return b;
    }
    ```

??? pythontutor "可视化运行"

    <div style="height: 513px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp_comp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20a,%20b%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20a,%20b%20%3D%20b,%20min%28a,%20b%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20b%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp_comp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
    <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp_comp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20a,%20b%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20a,%20b%20%3D%20b,%20min%28a,%20b%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20b%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp_comp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a></div>

## 14.2.2 &nbsp; 无后效性

无后效性是动态规划能够有效解决问题的重要特性之一，其定义为：**给定一个确定的状态，它的未来发展只与当前状态有关，而与过去经历的所有状态无关**。

以爬楼梯问题为例，给定状态 $i$ ，它会发展出状态 $i+1$ 和状态 $i+2$ ，分别对应跳 $1$ 步和跳 $2$ 步。在做出这两种选择时，我们无须考虑状态 $i$ 之前的状态，它们对状态 $i$ 的未来没有影响。

然而，如果我们给爬楼梯问题添加一个约束，情况就不一样了。

!!! question "带约束爬楼梯"

    给定一个共有 $n$ 阶的楼梯，你每步可以上 $1$ 阶或者 $2$ 阶，**但不能连续两轮跳 $1$ 阶**，请问有多少种方案可以爬到楼顶？

如图 14-8 所示，爬上第 $3$ 阶仅剩 $2$ 种可行方案，其中连续三次跳 $1$ 阶的方案不满足约束条件，因此被舍弃。

![带约束爬到第 3 阶的方案数量](dp_problem_features.assets/climbing_stairs_constraint_example.png){ class="animation-figure" }

<p align="center"> 图 14-8 &nbsp; 带约束爬到第 3 阶的方案数量 </p>

在该问题中，如果上一轮是跳 $1$ 阶上来的，那么下一轮就必须跳 $2$ 阶。这意味着，**下一步选择不能由当前状态（当前所在楼梯阶数）独立决定，还和前一个状态（上一轮所在楼梯阶数）有关**。

不难发现，此问题已不满足无后效性，状态转移方程 $dp[i] = dp[i-1] + dp[i-2]$ 也失效了，因为 $dp[i-1]$ 代表本轮跳 $1$ 阶，但其中包含了许多“上一轮是跳 $1$ 阶上来的”方案，而为了满足约束，我们就不能将 $dp[i-1]$ 直接计入 $dp[i]$ 中。

为此，我们需要扩展状态定义：**状态 $[i, j]$ 表示处在第 $i$ 阶并且上一轮跳了 $j$ 阶**，其中 $j \in \{1, 2\}$ 。此状态定义有效地区分了上一轮跳了 $1$ 阶还是 $2$ 阶，我们可以据此判断当前状态是从何而来的。

- 当上一轮跳了 $1$ 阶时，上上一轮只能选择跳 $2$ 阶，即 $dp[i, 1]$ 只能从 $dp[i-1, 2]$ 转移过来。
- 当上一轮跳了 $2$ 阶时，上上一轮可选择跳 $1$ 阶或跳 $2$ 阶，即 $dp[i, 2]$ 可以从 $dp[i-2, 1]$ 或 $dp[i-2, 2]$ 转移过来。

如图 14-9 所示，在该定义下，$dp[i, j]$ 表示状态 $[i, j]$ 对应的方案数。此时状态转移方程为：

$$
\begin{cases}
dp[i, 1] = dp[i-1, 2] \\
dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
\end{cases}
$$

![考虑约束下的递推关系](dp_problem_features.assets/climbing_stairs_constraint_state_transfer.png){ class="animation-figure" }

<p align="center"> 图 14-9 &nbsp; 考虑约束下的递推关系 </p>

最终，返回 $dp[n, 1] + dp[n, 2]$ 即可，两者之和代表爬到第 $n$ 阶的方案总数：

=== "Python"

    ```python title="climbing_stairs_constraint_dp.py"
    def climbing_stairs_constraint_dp(n: int) -> int:
        """带约束爬楼梯：动态规划"""
        if n == 1 or n == 2:
            return 1
        # 初始化 dp 表，用于存储子问题的解
        dp = [[0] * 3 for _ in range(n + 1)]
        # 初始状态：预设最小子问题的解
        dp[1][1], dp[1][2] = 1, 0
        dp[2][1], dp[2][2] = 0, 1
        # 状态转移：从较小子问题逐步求解较大子问题
        for i in range(3, n + 1):
            dp[i][1] = dp[i - 1][2]
            dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
        return dp[n][1] + dp[n][2]
    ```

=== "C++"

    ```cpp title="climbing_stairs_constraint_dp.cpp"
    /* 带约束爬楼梯：动态规划 */
    int climbingStairsConstraintDP(int n) {
        if (n == 1 || n == 2) {
            return 1;
        }
        // 初始化 dp 表，用于存储子问题的解
        vector<vector<int>> dp(n + 1, vector<int>(3, 0));
        // 初始状态：预设最小子问题的解
        dp[1][1] = 1;
        dp[1][2] = 0;
        dp[2][1] = 0;
        dp[2][2] = 1;
        // 状态转移：从较小子问题逐步求解较大子问题
        for (int i = 3; i <= n; i++) {
            dp[i][1] = dp[i - 1][2];
            dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
        }
        return dp[n][1] + dp[n][2];
    }
    ```

=== "Java"

    ```java title="climbing_stairs_constraint_dp.java"
    /* 带约束爬楼梯：动态规划 */
    int climbingStairsConstraintDP(int n) {
        if (n == 1 || n == 2) {
            return 1;
        }
        // 初始化 dp 表，用于存储子问题的解
        int[][] dp = new int[n + 1][3];
        // 初始状态：预设最小子问题的解
        dp[1][1] = 1;
        dp[1][2] = 0;
        dp[2][1] = 0;
        dp[2][2] = 1;
        // 状态转移：从较小子问题逐步求解较大子问题
        for (int i = 3; i <= n; i++) {
            dp[i][1] = dp[i - 1][2];
            dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
        }
        return dp[n][1] + dp[n][2];
    }
    ```

=== "C#"

    ```csharp title="climbing_stairs_constraint_dp.cs"
    /* 带约束爬楼梯：动态规划 */
    int ClimbingStairsConstraintDP(int n) {
        if (n == 1 || n == 2) {
            return 1;
        }
        // 初始化 dp 表，用于存储子问题的解
        int[,] dp = new int[n + 1, 3];
        // 初始状态：预设最小子问题的解
        dp[1, 1] = 1;
        dp[1, 2] = 0;
        dp[2, 1] = 0;
        dp[2, 2] = 1;
        // 状态转移：从较小子问题逐步求解较大子问题
        for (int i = 3; i <= n; i++) {
            dp[i, 1] = dp[i - 1, 2];
            dp[i, 2] = dp[i - 2, 1] + dp[i - 2, 2];
        }
        return dp[n, 1] + dp[n, 2];
    }
    ```

=== "Go"

    ```go title="climbing_stairs_constraint_dp.go"
    /* 带约束爬楼梯：动态规划 */
    func climbingStairsConstraintDP(n int) int {
        if n == 1 || n == 2 {
            return 1
        }
        // 初始化 dp 表，用于存储子问题的解
        dp := make([][3]int, n+1)
        // 初始状态：预设最小子问题的解
        dp[1][1] = 1
        dp[1][2] = 0
        dp[2][1] = 0
        dp[2][2] = 1
        // 状态转移：从较小子问题逐步求解较大子问题
        for i := 3; i <= n; i++ {
            dp[i][1] = dp[i-1][2]
            dp[i][2] = dp[i-2][1] + dp[i-2][2]
        }
        return dp[n][1] + dp[n][2]
    }
    ```

=== "Swift"

    ```swift title="climbing_stairs_constraint_dp.swift"
    /* 带约束爬楼梯：动态规划 */
    func climbingStairsConstraintDP(n: Int) -> Int {
        if n == 1 || n == 2 {
            return 1
        }
        // 初始化 dp 表，用于存储子问题的解
        var dp = Array(repeating: Array(repeating: 0, count: 3), count: n + 1)
        // 初始状态：预设最小子问题的解
        dp[1][1] = 1
        dp[1][2] = 0
        dp[2][1] = 0
        dp[2][2] = 1
        // 状态转移：从较小子问题逐步求解较大子问题
        for i in 3 ... n {
            dp[i][1] = dp[i - 1][2]
            dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
        }
        return dp[n][1] + dp[n][2]
    }
    ```

=== "JS"

    ```javascript title="climbing_stairs_constraint_dp.js"
    /* 带约束爬楼梯：动态规划 */
    function climbingStairsConstraintDP(n) {
        if (n === 1 || n === 2) {
            return 1;
        }
        // 初始化 dp 表，用于存储子问题的解
        const dp = Array.from(new Array(n + 1), () => new Array(3));
        // 初始状态：预设最小子问题的解
        dp[1][1] = 1;
        dp[1][2] = 0;
        dp[2][1] = 0;
        dp[2][2] = 1;
        // 状态转移：从较小子问题逐步求解较大子问题
        for (let i = 3; i <= n; i++) {
            dp[i][1] = dp[i - 1][2];
            dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
        }
        return dp[n][1] + dp[n][2];
    }
    ```

=== "TS"

    ```typescript title="climbing_stairs_constraint_dp.ts"
    /* 带约束爬楼梯：动态规划 */
    function climbingStairsConstraintDP(n: number): number {
        if (n === 1 || n === 2) {
            return 1;
        }
        // 初始化 dp 表，用于存储子问题的解
        const dp = Array.from({ length: n + 1 }, () => new Array(3));
        // 初始状态：预设最小子问题的解
        dp[1][1] = 1;
        dp[1][2] = 0;
        dp[2][1] = 0;
        dp[2][2] = 1;
        // 状态转移：从较小子问题逐步求解较大子问题
        for (let i = 3; i <= n; i++) {
            dp[i][1] = dp[i - 1][2];
            dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
        }
        return dp[n][1] + dp[n][2];
    }
    ```

=== "Dart"

    ```dart title="climbing_stairs_constraint_dp.dart"
    /* 带约束爬楼梯：动态规划 */
    int climbingStairsConstraintDP(int n) {
      if (n == 1 || n == 2) {
        return 1;
      }
      // 初始化 dp 表，用于存储子问题的解
      List<List<int>> dp = List.generate(n + 1, (index) => List.filled(3, 0));
      // 初始状态：预设最小子问题的解
      dp[1][1] = 1;
      dp[1][2] = 0;
      dp[2][1] = 0;
      dp[2][2] = 1;
      // 状态转移：从较小子问题逐步求解较大子问题
      for (int i = 3; i <= n; i++) {
        dp[i][1] = dp[i - 1][2];
        dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
      }
      return dp[n][1] + dp[n][2];
    }
    ```

=== "Rust"

    ```rust title="climbing_stairs_constraint_dp.rs"
    /* 带约束爬楼梯：动态规划 */
    fn climbing_stairs_constraint_dp(n: usize) -> i32 {
        if n == 1 || n == 2 {
            return 1;
        };
        // 初始化 dp 表，用于存储子问题的解
        let mut dp = vec![vec![-1; 3]; n + 1];
        // 初始状态：预设最小子问题的解
        dp[1][1] = 1;
        dp[1][2] = 0;
        dp[2][1] = 0;
        dp[2][2] = 1;
        // 状态转移：从较小子问题逐步求解较大子问题
        for i in 3..=n {
            dp[i][1] = dp[i - 1][2];
            dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
        }
        dp[n][1] + dp[n][2]
    }
    ```

=== "C"

    ```c title="climbing_stairs_constraint_dp.c"
    /* 带约束爬楼梯：动态规划 */
    int climbingStairsConstraintDP(int n) {
        if (n == 1 || n == 2) {
            return 1;
        }
        // 初始化 dp 表，用于存储子问题的解
        int **dp = malloc((n + 1) * sizeof(int *));
        for (int i = 0; i <= n; i++) {
            dp[i] = calloc(3, sizeof(int));
        }
        // 初始状态：预设最小子问题的解
        dp[1][1] = 1;
        dp[1][2] = 0;
        dp[2][1] = 0;
        dp[2][2] = 1;
        // 状态转移：从较小子问题逐步求解较大子问题
        for (int i = 3; i <= n; i++) {
            dp[i][1] = dp[i - 1][2];
            dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
        }
        int res = dp[n][1] + dp[n][2];
        // 释放内存
        for (int i = 0; i <= n; i++) {
            free(dp[i]);
        }
        free(dp);
        return res;
    }
    ```

=== "Kotlin"

    ```kotlin title="climbing_stairs_constraint_dp.kt"
    /* 带约束爬楼梯：动态规划 */
    fun climbingStairsConstraintDP(n: Int): Int {
        if (n == 1 || n == 2) {
            return 1
        }
        // 初始化 dp 表，用于存储子问题的解
        val dp = Array(n + 1) { IntArray(3) }
        // 初始状态：预设最小子问题的解
        dp[1][1] = 1
        dp[1][2] = 0
        dp[2][1] = 0
        dp[2][2] = 1
        // 状态转移：从较小子问题逐步求解较大子问题
        for (i in 3..n) {
            dp[i][1] = dp[i - 1][2]
            dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
        }
        return dp[n][1] + dp[n][2]
    }
    ```

=== "Ruby"

    ```ruby title="climbing_stairs_constraint_dp.rb"
    [class]{}-[func]{climbing_stairs_constraint_dp}
    ```

=== "Zig"

    ```zig title="climbing_stairs_constraint_dp.zig"
    // 带约束爬楼梯：动态规划
    fn climbingStairsConstraintDP(comptime n: usize) i32 {
        if (n == 1 or n == 2) {
            return 1;
        }
        // 初始化 dp 表，用于存储子问题的解
        var dp = [_][3]i32{ [_]i32{ -1, -1, -1 } } ** (n + 1);
        // 初始状态：预设最小子问题的解
        dp[1][1] = 1;
        dp[1][2] = 0;
        dp[2][1] = 0;
        dp[2][2] = 1;
        // 状态转移：从较小子问题逐步求解较大子问题
        for (3..n + 1) |i| {
            dp[i][1] = dp[i - 1][2];
            dp[i][2] = dp[i - 2][1] + dp[i - 2][2];
        }
        return dp[n][1] + dp[n][2];
    }
    ```

??? pythontutor "可视化运行"

    <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_constraint_dp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B8%A6%E7%BA%A6%E6%9D%9F%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%203%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D%5B1%5D,%20dp%5B1%5D%5B2%5D%20%3D%201,%200%0A%20%20%20%20dp%5B2%5D%5B1%5D,%20dp%5B2%5D%5B2%5D%20%3D%200,%201%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B1%5D%20%3D%20dp%5Bi%20-%201%5D%5B2%5D%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B2%5D%20%3D%20dp%5Bi%20-%202%5D%5B1%5D%20%2B%20dp%5Bi%20-%202%5D%5B2%5D%0A%20%20%20%20return%20dp%5Bn%5D%5B1%5D%20%2B%20dp%5Bn%5D%5B2%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_constraint_dp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
    <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_constraint_dp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B8%A6%E7%BA%A6%E6%9D%9F%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%203%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D%5B1%5D,%20dp%5B1%5D%5B2%5D%20%3D%201,%200%0A%20%20%20%20dp%5B2%5D%5B1%5D,%20dp%5B2%5D%5B2%5D%20%3D%200,%201%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B1%5D%20%3D%20dp%5Bi%20-%201%5D%5B2%5D%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B2%5D%20%3D%20dp%5Bi%20-%202%5D%5B1%5D%20%2B%20dp%5Bi%20-%202%5D%5B2%5D%0A%20%20%20%20return%20dp%5Bn%5D%5B1%5D%20%2B%20dp%5Bn%5D%5B2%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_constraint_dp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a></div>

在上面的案例中，由于仅需多考虑前面一个状态，因此我们仍然可以通过扩展状态定义，使得问题重新满足无后效性。然而，某些问题具有非常严重的“有后效性”。

!!! question "爬楼梯与障碍生成"

    给定一个共有 $n$ 阶的楼梯，你每步可以上 $1$ 阶或者 $2$ 阶。**规定当爬到第 $i$ 阶时，系统自动会在第 $2i$ 阶上放上障碍物，之后所有轮都不允许跳到第 $2i$ 阶上**。例如，前两轮分别跳到了第 $2$、$3$ 阶上，则之后就不能跳到第 $4$、$6$ 阶上。请问有多少种方案可以爬到楼顶？

在这个问题中，下次跳跃依赖过去所有的状态，因为每一次跳跃都会在更高的阶梯上设置障碍，并影响未来的跳跃。对于这类问题，动态规划往往难以解决。

实际上，许多复杂的组合优化问题（例如旅行商问题）不满足无后效性。对于这类问题，我们通常会选择使用其他方法，例如启发式搜索、遗传算法、强化学习等，从而在有限时间内得到可用的局部最优解。
