Add the section of unbounded knapsack problem.

codes/python/chapter_dynamic_programming/coin_change.py

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+"""
+File: coin_change.py
+Created Time: 2023-07-10
+Author: Krahets (krahets@163.com)
+"""
+
+
+def coin_change_dp(coins: list[int], amt: int) -> int:
+    """零钱兑换：动态规划"""
+    n = len(coins)
+    MAX = amt + 1
+    # 初始化 dp 表
+    dp = [[0] * (amt + 1) for _ in range(n + 1)]
+    # 状态转移：首行首列
+    for a in range(1, amt + 1):
+        dp[0][a] = MAX
+    # 状态转移：其余行列
+    for i in range(1, n + 1):
+        for a in range(1, amt + 1):
+            if coins[i - 1] > a:
+                # 若超过背包容量，则不选硬币 i
+                dp[i][a] = dp[i - 1][a]
+            else:
+                # 不选和选硬币 i 这两种方案的较小值
+                dp[i][a] = min(dp[i - 1][a], dp[i][a - coins[i - 1]] + 1)
+    return dp[n][amt] if dp[n][amt] != MAX else -1
+
+
+def coin_change_dp_comp(coins: list[int], amt: int) -> int:
+    """零钱兑换：状态压缩后的动态规划"""
+    n = len(coins)
+    MAX = amt + 1
+    # 初始化 dp 表
+    dp = [MAX] * (amt + 1)
+    dp[0] = 0
+    # 状态转移
+    for i in range(1, n + 1):
+        # 正序遍历
+        for a in range(1, amt + 1):
+            if coins[i - 1] > a:
+                # 若超过背包容量，则不选硬币 i
+                dp[a] = dp[a]
+            else:
+                # 不选和选硬币 i 这两种方案的较小值
+                dp[a] = min(dp[a], dp[a - coins[i - 1]] + 1)
+    return dp[amt] if dp[amt] != MAX else -1
+
+
+"""Driver Code"""
+if __name__ == "__main__":
+    coins = [1, 2, 5]
+    amt = 4
+
+    # 动态规划
+    res = coin_change_dp(coins, amt)
+    print(f"凑到目标金额所需的最少硬币数量为 {res}")
+
+    # 状态压缩后的动态规划
+    res = coin_change_dp_comp(coins, amt)
+    print(f"凑到目标金额所需的最少硬币数量为 {res}")