硬币凑整问题——背包01

有n种不同面值的硬币，每种一个；试着用这些硬币凑出整数m
如果有多种方案，输出字典序最小的那个
eg.
input:
3 5
3 2 6
output: 2 3
input:
8 9
5 9 8 7 2 3 4 1
output:
1 3 5

• 说明一：这里的硬币逻辑上是价值和质量相同的物品w[i]=v[i],用m_w_v[]一个数组存放
• 说明二：字典序最小，先将面值由大到小排序，最后按记录结果倒序输出

#pragma once

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

const int MAXSIZE = 13;


class BAGS
{
    
    
public:
	BAGS() {
    
    
		init();
	}
	~BAGS() {
    
    }

	void init() {
    
    
		int n = 0,t=0;
		cin >> n >> m_bagCapcity;
		for (int i = 0; i < n; i++) {
    
    
			cin >> t;
			m_w_v.push_back(t);
		}
		sort(m_w_v.begin(), m_w_v.end(), [=](int i, int j)->bool {
    
     return i > j; });
		
		for (int i = 0; i < MAXSIZE; i++) {
    
    
			for (int j = 0; j < MAXSIZE; j++) {
    
    
				m_choice[i][j] = false;
			}
		}
	}

	void getMatchPlan() {
    
    
		vector<int> maxValue;
		for (int i = 0; i < 1+ m_bagCapcity; i++) {
    
    
			maxValue.push_back(0);
		}

		getChoiceTable(maxValue);

		if (maxValue[m_bagCapcity] != m_bagCapcity) {
    
    
			printf_s("No soluton.\n");
		}
		else {
    
    
			printPlan();
		}

	}
		
	void getChoiceTable(vector<int>& maxValue) {
    
    
		for (int i = 0; i < m_w_v.size(); i++) {
    
    
			for (int cap = m_bagCapcity; cap >= m_w_v[i]; cap--) {
    
    //attention.
				if (maxValue[cap] <= maxValue[cap - m_w_v[i]] + m_w_v[i]) {
    
    
					maxValue[cap] = maxValue[cap - m_w_v[i]] + m_w_v[i];
					m_choice[i][cap] = true;
				}
				else {
    
    
					m_choice[i][cap] = false;
				}
				//cout << cap << " " << maxValue[cap]<<endl;
			}
		}
	}

	void printPlan() {
    
    
		int n = m_w_v.size()-1;
		int cap = m_bagCapcity;
		while (n >= 0) {
    
    
			if (m_choice[n][cap]) {
    
    
				cout<<m_w_v[n]<<" ";
				cap -= m_w_v[n];
			}
			n--;
		}
		cout << endl;
	}

	
private:
	vector<int> m_w_v;//weight=value
	int m_bagCapcity;
	bool m_choice[MAXSIZE][MAXSIZE];//choice[i][c] 容量为c时，第i件物品是否选中
};

/*
8 9
5 9 8 7 2 3 4 1
*/

void testForBag01() {
    
    
	BAGS test;
	test.getMatchPlan();
}

1. 要小心下标和表格一致，例如取不取边界值，起点是谁？这里硬币种类从零开始到n-1,而容量[0,capacity]
2. 注意更新条件是capacity>当前某个值的权重，而不是大于零