算法学习：用回溯解决01背包问题_利用回溯法的递归框架,实现 01 背包问题,(问题设定:背包的限定重 量为 6,物品数量

1.设限重W,装入总价值cp,总重量cw.

2.需要设置布尔数组来储存物品的存放情况，0为无，1为有。

3.最优解bestp,最优方法bestb[n]

4.方法数组为bool型，输入数量为int型，重量和价值为double型。

注意：运行条件，终止条件

#include<iostream>
using namespace std;
 
typedef struct sub {
	double weight;
	double value;
}S;
 
double bound(int i, S* s, double cp, int n);
void Backrack(int i, bool* b, S* s, int n, double cp, double cw, double W, bool* bestb);
 
double bestp = 0;
 
int main() {
	double cp = 0, cw = 0;
	int n, W;
	cout << "输入数量和限重" << endl;
	cin >> n >> W;
	S* s = new S[n];
	bool* b = new bool[n];
	bool* bestb = new bool[n];
	
 
	cout << "输入价值，重量" << endl;
	for (int i = 0; i < n; i++) {
		cin >> s[i].value >> s[i].weight;
	}
	Backrack(0, b, s, n, cp, cw, W,bestb);
	cout << bestp<<endl;
	for (int i = 0; i < n; i++) {
		cout << bestb[i] << " ";
	}
}
 
double bound(int i, S* s, double cp, int n) { 
	double rp=0;
	for (int j = i; j < n; j++) {
		rp += s[j].value;
	}
	return rp+cp;
}
 
 
void Backrack(int i, bool *b,S *s,int n,double cp,double cw,double W,bool *bestb) {
	if (i > n-1) {
		int c=0;
 
		if (cp > bestp) {
			bestp = cp;
			for (int i = 0; i < n; i++) {
				bestb[i] = b[i];
			}
		}
		return;
	}
	if (cw + s[i].weight <= W) {
		cp += s[i].value;
		cw += s[i].weight;
		b[i] = 1;
		Backrack(i + 1, b, s, n, cp, cw, W,bestb);
		cp -= s[i].value;
		cw -= s[i].weight;
	}
	if (bound(i + 1,s,cp,n) > bestp) {
		b[i] = 0;
		Backrack(i + 1, b, s, n, cp, cw, W,bestb);
	}
}