C# 动态规划(Dynamic Programming)-1

动态规划算法：

     动态规划算法通常基于一个递推公式及一个或多个初始状态。当前子问题的解将由上一次子问题的解推出。动态规划和分治法相似，都是通过组合子问题的解来求解原问题。分治法将问题划分成互不相交的子问题，递归求解子问题，再将他们的解组合起来，求出原问题的解。与之相反，动态规划应用于子问题重叠的情况，即不同的子问题具有公共的子子问题。在这种情况下，分治算法会做出许多不必要的工作，它会反复的求解那些公共子问题。而动态规划算法对每个子子问题只求解一次，将结果保存到表格中，从而无需每次求解一个子子问题都要重新计算。

 钢条切割问题：

 常规递归方法：

internal class Program
    {
        static void Main(string[] args)
        {
            int n = 5;//要切割的钢条长度
            int[] price = { 0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 30 };//数值=价格，索引=钢条长度
            Console.WriteLine(MaxPrice(0,price));
            Console.WriteLine(MaxPrice(1,price));
            Console.WriteLine(MaxPrice(2,price));
            Console.WriteLine(MaxPrice(3,price));
            Console.WriteLine(MaxPrice(4,price));
            Console.WriteLine(MaxPrice(5,price));
            Console.WriteLine(MaxPrice(6,price));
            Console.WriteLine(MaxPrice(7,price));
            Console.WriteLine(MaxPrice(8,price));
            Console.WriteLine(MaxPrice(9,price));
            Console.WriteLine(MaxPrice(10,price));
        }
        public static int MaxPrice(int n, int[] p)
        {
            if (n == 0) return 0;
            int tempMaxPrice = 0;
            for(int i = 1; i < n+1; i++)
            {
                int maxPrice = p[i]+MaxPrice(n-i, p);
                if(maxPrice > tempMaxPrice)
                {
                    tempMaxPrice = maxPrice;
                }
            }
            return tempMaxPrice;
        }
    }

 动态规划-自顶向下法：

internal class Program
    {
        static void Main(string[] args)
        {
            int n = 5;//要切割的钢条长度
            int[] memory = new int[11];
            int[] price = { 0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 30 };//数值=价格，索引=钢条长度
            Console.WriteLine(MaxPrice(0, price, memory));
            Console.WriteLine(MaxPrice(1, price, memory));
            Console.WriteLine(MaxPrice(2, price, memory));
            Console.WriteLine(MaxPrice(3, price, memory));
            Console.WriteLine(MaxPrice(4, price, memory));
            Console.WriteLine(MaxPrice(5, price, memory));
            Console.WriteLine(MaxPrice(6, price, memory));
            Console.WriteLine(MaxPrice(7, price, memory));
            Console.WriteLine(MaxPrice(8, price, memory));
            Console.WriteLine(MaxPrice(9, price, memory));
            Console.WriteLine(MaxPrice(10, price, memory));
        }
        public static int MaxPrice(int n, int[] p,int[] memory)
        {
            if (n == 0) return 0;
            if (memory[n] != 0)
            {
                return memory[n];
            }
            int tempMaxPrice = 0;
            for (int i = 1; i < n + 1; i++)
            {
                int maxPrice = p[i] + MaxPrice(n - i, p,memory);
                if (maxPrice > tempMaxPrice)
                {
                    tempMaxPrice = maxPrice;
                }
            }
            memory[n] = tempMaxPrice;
            return tempMaxPrice;
        }
    }

动态规划-自底向上法：

internal class Program
    {
        static void Main(string[] args)
        {
            int[] memory = new int[11];
            int[] price = { 0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 30 };//数值=价格，索引=钢条长度
            Console.WriteLine(MaxPrice(0, price, memory));
            Console.WriteLine(MaxPrice(1, price, memory));
            Console.WriteLine(MaxPrice(2, price, memory));
            Console.WriteLine(MaxPrice(3, price, memory));
            Console.WriteLine(MaxPrice(4, price, memory));
            Console.WriteLine(MaxPrice(5, price, memory));
            Console.WriteLine(MaxPrice(6, price, memory));
            Console.WriteLine(MaxPrice(7, price, memory));
            Console.WriteLine(MaxPrice(8, price, memory));
            Console.WriteLine(MaxPrice(9, price, memory));
            Console.WriteLine(MaxPrice(10, price, memory));
        }
        public static int MaxPrice(int n, int[] p, int[] memory)
        {
            for (int i = 1; i < n + 1; i++) 
            {
                //钢条长度为i时最大收益
                int tempMaxPrice = 0;
                for(int j = 1; j < i+1; j++)
                {
                    int maxPrice = p[j] + memory[i-j];
                    if (maxPrice > tempMaxPrice)
                    {
                        tempMaxPrice = maxPrice;
                    }
                }
                memory[i] = tempMaxPrice;
            }
            return memory[n];
        }
    }

 结果：