01背包问题VS完全背包问题

01背包问题：
每个物品只能选择一次，给定背包的最大承重量total和每个物品i的重量weight[i]和价值value[i],求背包可以装下的最大价值
 

#include<iostream>
#include<vector>
using namespace std;

int main()
{
	int num, total;
	cin >> num >> total;
	vector<int> weight(num+1);
	vector<int> value(num+1);
	for (int i = 1; i <= num; i++)
	{
		cin >> weight[i] >> value[i];
	}
	vector<vector<int>> f(num+1, vector<int>(total+1, 0));
	for (int i = 1; i <= num; i++)
	{
		for (int j = total; j >= 1; j--)
		{
			if (j >= weight[i])
			{
				f[i][j] = max(f[i - 1][j], f[i - 1][j - weight[i]] + value[i]);
			}
		}
	}
	cout << f[num][total] << endl;
	return 0;
}

优化之后：

#include<iostream>
#include<vector>
using namespace std;

int main()
{
	int num, total;
	cin >> num >> total;
	vector<int> weight(num+1);
	vector<int> value(num+1);
	for (int i = 1; i <= num; i++)
	{
		cin >> weight[i] >> value[i];
	}
	vector<int> f(total+1, 0);
	for (int i = 1; i <= num; i++)
	{
		for (int j = total; j >= 1; j--)
		{
			if (j >= weight[i])
			{
				f[j] = max(f[j], f[j - weight[i]] + value[i]);
			}
		}
	}
	cout << f[total] << endl;
	return 0;
}

完全背包问题：

基本题意和01背包问题一样，只不过在选择物品过程中每个物品可以选择多次

#include<iostream>
#include<vector>
using namespace std;

int main()
{
	int num, total;
	cin >> num >> total;
	vector<int> weight(num+1);
	vector<int> value(num+1);
	vector<vector<int>> f(num + 1, vector<int>(total + 1, 0));
	for (int i = 1; i <= num; i++)
	{
		cin >> weight[i] >> value[i];
	}
	for (int i = 1; i <= num; i++)
	{
		for (int j = 1; j <= total; j++)
		{
			if (j >= weight[i])
			{
				f[i][j] = max(f[i - 1][j], f[i][j - weight[i]] + value[i]);
			}
		}
	}
	cout << f[num][total] << endl;
	return 0;
}

优化之后：

#include<iostream>
#include<vector>
using namespace std;

int main()
{
	int num, total;
	cin >> num >> total;
	vector<int> weight(num+1);
	vector<int> value(num+1);
	vector<int> f(total+1);
	for (int i = 1; i <= num; i++)
	{
		cin >> weight[i] >> value [i];
	}
	for (int i = 1; i <= num; i++)
	{
		for (int j=1;j<=total;j++)
		{
			if (j>=weight[i])
			{
				f[j] = max(f[j], f[j - weight[i]] + value[i]);
			}
		}
	}
	cout << f[total] << endl;
	return 0;
}

区别1:

01背包问题的状态转移矩阵为：

f[i][j]=max(f[i-1][j],f[i-1][j-weight[i]]+value[i])

f[j]=max(f[j],f[j-weight[i]]+value[i])

完全背包的状态转移矩阵为：

f[i][j]=max(f[i-1],f[i][j-weight[i]]+value[i])

f[j]=max(f[j],f[j-weight[i]]+value[i])

区别2:

01背包矩阵遍历剩余承重量时倒序，完全背包遍历时正序。