Greedy method (non-0-1 knapsack problem)

给定背包容量50Kg，物品信息
物品1，重量10Kg，价值60元
物品2，重量20Kg，价值100元
物品3，重量30Kg，价值120元
注意：可以部分装入

Pictures and topics are transferred from https://blog.csdn.net/songshimvp1/article/details/52350542

#pragma once

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<stack>
using namespace std;

const int MAXSIZE = 3;


class BAGS
{
    
    
public:
	BAGS(){
    
    }
	BAGS(int i) {
    
    
		if (i == 1) {
    
    
			init();
		}
	}
	~BAGS(){
    
    }

	void init() {
    
    
		m_weight[0] = 20;
		m_value[0] = 100;
		m_VEachW[0] = 5;
		m_weight[1] = 30;
		m_value[1] = 120;
		m_VEachW[1] = 4;
		m_weight[2] = 10;
		m_value[2] = 60;
		m_VEachW[2] = 6;

		m_bagCapcity = 50;
	}

	void getMaxValue() {
    
    
		vector<int> pos;
		for (int i = 0; i < MAXSIZE; i++) {
    
    
			pos.push_back(i);
		}

		sortPosByVeachW(pos);
		//for(int i:pos) cout << m_VEachW[i] << endl;

		int maxValue = encloseByPos(pos);
		cout << maxValue << endl;
	}

	void sortPosByVeachW(vector<int>& pos) {
    
    
		for (int i = 0; i < MAXSIZE; i++) {
    
    
			bool hasUpdate = false;
			for (int j = 0; j+1 < MAXSIZE; j++) {
    
    
				int pos1 = pos[j], pos2 = pos[j + 1];//important!!
				if (m_VEachW[pos1] < m_VEachW[pos2]) {
    
    
					swap(pos[j], pos[j+1]);
					hasUpdate = true;
				}
			}
			if (!hasUpdate) break;
		}
	}

	int encloseByPos(vector<int> &pos) {
    
    
		int maxValue = 0;
		int i = 0;
		int curPos = pos.at(i);
		int residualCapcity = m_bagCapcity;

		while (residualCapcity != 0) {
    
    
			if (residualCapcity >= m_weight[curPos]) {
    
    
				maxValue += m_value[curPos];
				residualCapcity -= m_weight[curPos];
				i++;
				if(i<MAXSIZE)
				curPos = pos[i];
			}
			else {
    
    
				maxValue += residualCapcity * m_VEachW[curPos];
				break;
			}
		}
		
		return maxValue;
	}

private:
	int m_weight[MAXSIZE], m_value[MAXSIZE],m_VEachW[MAXSIZE];
	int m_bagCapcity;
};

void testForBag() {
    
    
	BAGS test(1);
	test.getMaxValue();
}

要注意的：
1，按 元/kg 做关联排序，即只对位置排序，小心交换的位置不是直接下标，而是pos数组中的数据
for (int j = 0; j+1 < MAXSIZE; j++) {
    
    
	int pos1 = pos[j], pos2 = pos[j + 1];//important!!
	if (m_VEachW[pos1] < m_VEachW[pos2]) {
    
    
				swap(pos[j], pos[j+1]);
				hasUpdate = true；
	}
2，函数的返回值尽量为空，大可以把返回值作为引用，如下
void sortPosByVeachW(vector<int>& pos)；//推荐使用一个输入，返回void
vector<int>& sortPosByVeachW()；
3，out_of_range：
      i++;
	  if(i<MAXSIZE) curPos = pos[i];
4，贪心法：每次装入的都是单位价值最高的