背包问题合集（动态规划方法）

1. 01背包（一维数组优化）

   #include<iostream>
   using namespace std;
   const int N = 110;%物品数量 
   int v[N], w[N], f[N];%分别是物品的体积，价值，当前状态的最大价值 
   int main()
   {
       int n = 0, m = 0;
       cin >> n >> m;
       for (int i = 1; i <= n; i++)
           cin >> v[i] >> w[i];
       for (int i = 1; i <= n; i++)
           for (int j = m; j >= v[i]; j--) //从大到小
           {
               f[j] = max(f[j], f[j - v[i]] + w[i]);    
           }
           cout << f[m];
       return 0;
   }

   3件物品，有体积和价值，背包最大体积为10

2. 完全背包问题（二维数组优化）

   #include<iostream>
   using namespace std;
   const int N = 100;%物品数量 
   const int N = 1e4;%背包容量 
   int v[N], w[N], f[N][N];
   int main()
   {
   	int m, n;
   	cin >> m >> n;
   	for (int i = 1; i <= n; i++)
   	{
   		cin >> v[i] >> w[i];%输入物品体积和价值 
   	}
   	for (int i = 1; i <= n; i++)
   		for (int j = 1; j <= m; j++)
   		{
   			f[i][j] = f[i - 1][j];
   			if (j >= v[i])
   			{
   				f[i][j] = max(f[i][j], f[i][j - v[i]] + w[i]);
   			
   			}
   		}
   	cout << f[n][m];%前n个物品，容量为m的最大价值 
   }

   

3. 多组背包问题（先分组，在01背包）

   #include<iostream>
   using namespace std;
   const int N = 1010;
   int m, n;
   int f[N], v[N][N], w[N][N], s[N], y[N];
   int main()
   {
   	int cnt = 0, M = 0, x = 0, MAX = 0;
   	cin >> m >> n;
   	for (int i = 1; i <= n; i++)
   	{
   		int a, b, c;
   		cin >> a >> b >> c;
   		int j = 1;
   			while (v[c][j] != 0)
   			{
   				j++;
   			}
   			v[c][j] = a;
   			w[c][j] = b;
   			y[c]++;
   			j = 1;
   		MAX = max(MAX, c);
   	}
   	for (int i = 1; i <= MAX; i++)
   	{
   		for (int j = m; j >=0; j--)
   		{
   			for (int k = 1; k <= y[i]; k++)
   			{
   				if (j >= v[i][k])
   				{
   					f[j] = max(f[j], f[j - v[i][k]] + w[i][k]);
   				}
   			}
   			cout << f[j] << endl;
   		}
   	}
   	return 0;
   }
   /*
   for (int i = 1; i <= MAX; i++)
   	{
   		for (int j = 1; j <=m; j++)
   		{
   			for (int k = 1; k <= y[i]; k++)
   			{
   				if (j >= v[i][k])
   				{
   					f[i][j] = f[i - 1][j];
   					f[i][j]= max(f[i][j], f[i - 1][j - v[i][k]] + w[i][k]);
   				}
   			}
   		}
   	}
   	cout << f[n][m];
   */