动态规划之钢条切割

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自顶向下递归实现(Recursive top-down implementation)
程序CUT-ROD对等式(14.2)进行了实现，伪代码如下：

CUT-ROD(p, n)if n == 0return 0q = -∞for i = 1 to nq = max{q, p[i] + CUT-ROD(p, n - i)}return q

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上面解决中重复对一个子结构问题重复求解了，我们可以把这个过程记录下来

使用动态规划求解最优钢条切割(Using dynamic programming for optimal rod cutting)

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MEMOIZED-CUT-ROD(p, n)let r[0 : n] be a new array    // will remember solution values in rfor i = 0 to nr[i] = -∞return MEMOIZED-CUT-ROD-AUX(p, n, r)MEMOIZED-CUT-ROD-AUX(p, n, r)if r[n] ≥ 0    // already have a solution for length n ?return r[n]if n == 0q = 0else q = -∞for i = 1 to n    // i is the position of the first cutq = max{q, p[i] + MEMOIZED-CUT-ROD-AUX(p, n - i, r)}r[n] = q    // remember the solution value for length nreturn q

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代码

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>class Solution {
public:Solution(int x):len(x){}int fun() {for (int i = 0; i <= len; i++) {memory[i] = 0;  // 初始化}for (int i = 1; i <= len; i++){int q = min;for (int k = 1; k <= i; k++){q =std::max( q , price[k] + memory[i - k]);   // 每次切割一次，这是核心record[i] = k;}memory[i] = q;}return memory[len];}private:int price[11] = {0,1,5,8,9,10,17,17,20,24,30};  // 假定数据较小,对应价格int len;int memory[11];   // 记录 int record[11];   // 在哪里切割int min = -1;
};int main()
{Solution a(4);std::cout <<a.fun() << std::endl;
}