使用回溯法解決0-1揹包問題

#include<iostream>
using namespace std;


int n, c, bestp;		//物品的個數，揹包的容量，最大價值
int p[80], w[80], x[80], bestx[80];		//物品的價值，物品的重量，x[i]暫存物品的選中情況,物品的選中情況

void Backtrack(int i, int cp, int cw)
{ //cw當前包內物品重量，cp當前包內物品價值
	int j;
	if (i>n)//回溯結束
	{
		if (cp>bestp)
		{
			bestp = cp;
			for
 
 (i = 0; i <= n; i++) bestx[i] = x[i];
		}
	}
	else
	for (j = 0; j <= 1; j++)
	{
		x[i] = j;
		if (cw + x[i] * w[i] <= c)
		{
			cw += w[i] * x[i];
			cp += p[i] * x[i];
			Backtrack(i + 1, cp, cw);
			cw -= w[i] * x[i];
			cp -= p[i] * x[i];
		}
	}
}

int _tmain(int argc, _TCHAR* argv[])
{
	bestp =
 
 0;
	cout << "輸入揹包最大容量,物品個數,物品的重量,物品的價值\n";
	cin >> c;
	cin >> n;
	for (int i = 1; i <= n; i++)cin >> w[i];
	for (int i = 1; i <= n; i++)cin >> p[i];
	Backtrack(1, 0, 0);
	cout << "最大價值為:" << bestp;
	printf("\n被選中的物品依次是\n");
	for (int i = 1; i <= n; i++)
 
if (bestx[i])cout << "第" << i << "個物品" << endl;
	system("pause>nul");
	return 0;
}