C++实现——动态规划法（切割钢条问题）-创新互联

假定公司出售一段长度为i英寸的钢条的利润为Pi，那么现有一根N英寸的钢条该如何切割才能利润大，大为多少。

#include#includeusing namespace std;

void mycut_roin(const int p[], int length, int n, int optim_income[], int optim_solution[]);
void show_array(int arr[], int n);

int main()
{const int length = 10;
	int p[length+1] = {0,1,5,8,9,10,17,17,20,24,30 };
	int n;
	cin >>n;
	int* optim_income(new int[n + 1]);
	int* optim_solution(new int[n + 1]);
	mycut_roin(p, length, n, optim_income, optim_solution);
	cout<< "The optimal income: "<< optim_income[n]<< endl;
	
	while (n >0)
	{cout<< optim_solution[n]<< ' ';
		n -= optim_solution[n];
	}
	delete[] optim_income;
	delete[] optim_solution;
	return 0;
}

void mycut_roin(const int p[], int length, int n, int optim_income[], int optim_solution[])
{optim_income[0] = optim_solution[0] = 0;
	for (int i = 1; i<= n; i++)
	{int max_income = INT_MIN;
		for (int j = 1; j<= i; j++)
		{	int pval = (j >= length) ? p[length] : p[j];
			if (pval + optim_income[i - j] >max_income)
			{		max_income = pval + optim_income[i - j];
				optim_solution[i] = j;
			}
		}
		optim_income[i] = max_income;
	}
	show_array(optim_income, n + 1);
	show_array(optim_solution, n + 1);
	
}

void show_array(int arr[], int n)
{for (int i = 0; i< n; i++)
	{cout.width(4);
		cout<< arr[i]<< ' ';
	}
	cout<< endl;
}

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