AcWing算法基础课---第五讲动态规划

背包问题

1. 01背包
   (1). 二维dp

   #include<iostream>
   
   using namespace std;
   
   const int N = 1010;
   int m, n;
   int V[N], W[N];
   int f[N][N]; //f[i][j]表示从i个物品中选，容量不超过j的最大价值
   
   int main()
   {
       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 = 0; j <= m; j ++) 
           {
               f[i][j] = f[i - 1][j]; //不包含i肯定存在
               if (j >= V[i]) f[i][j] = max(f[i][j], f[i - 1][j - V[i]] + W[i]); //包含i可能不存在
              
           }
       }
       
       cout << f[n][m];
       return 0;
   }
   

   (2). 压缩为一维dp

   #include<iostream>
   
   using namespace std;
   
   const int N = 1010;
   int m, n;
   int V[N], W[N];
   int f[N]; //f[i][j]表示从i个物品中选，容量不超过j的最大价值
   
   int main()
   {
       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 --) 
           {
               f[j] = max(f[j], f[j - V[i]] + W[i]); //包含i可能不存在
              
           }
       
       
       cout << f[m];
       return 0;
   }
   

2. 完全背包
   (1). 二维

   #include<iostream>
   
   using namespace std;
   
   const int N = 1010;
   
   int n, m;
   int v[N], w[N];
   int f[N][N];
   
   int main()
   {
       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 = 0; 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];
       return 0;
   }
   

   (2). 一维

   #include<iostream>
   
   using namespace std;
   
   const int N = 1010;
   
   int n, m;
   int v[N], w[N];
   int f[N];
   
   int main()
   {
       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 = v[i]; j <= m; j ++) 
               f[j] = max(f[j], f[j - v[i]] + w[i]);
           
           
       cout << f[m];
       return 0;
   }
   

3. 多重背包

   #include<iostream>
   
   using namespace std;
   
   const int N = 10010;
   
   int n, m;
   int v[N], w[N]; 
   int f[N]; 
   
   int main()
   {
       cin >> n >> m;
       
       int cnt = 0;
       for (int i = 1; i <= n; i ++) 
       {
           int a, b, s;
           cin >> a >> b >> s;
           int k = 1;
           
           //打包
           while (k <= s) 
           {
               cnt ++;
               v[cnt] = a * k;
               w[cnt] = b * k;
               s -= k;
               k *= 2;
           }
           
           if (s > 0) 
           {
               cnt ++;
               v[cnt] = a * s;
               w[cnt] = b * s;
           }
       }
       
       n = cnt;
       
       for (int i = 1; i <= n; i ++)
           for (int j = m; j >= v[i]; j --) 
               f[j] = max(f[j], f[j - v[i]] + w[i]);
               
       cout << f[m];
       
       return 0;
   }
   

4. 分组背包

   #include <iostream>
   
   using namespace std;
   
   const int N = 110;
   
   int n, m;
   int v[N][N], w[N][N], s[N], f[N];
   
   int main()
   {
       cin >> n >> m;
       
       for (int i = 1; i <= n; i ++) 
       {
           cin >> s[i];
           for (int j = 0; j < s[i]; j ++) 
               cin >> v[i][j] >> w[i][j];
       }
       
       for (int i = 1; i <= n; i ++) 
           for (int j = m; j >= 0; j --)
               for (int k = 0; k < s[i]; k ++)
                   if (v[i][k] <= j) //判断是否能装下！！！
                       f[j] = max(f[j], f[j - v[i][k]] + w[i][k]);
                   
       cout << f[m];
       
       return 0;
   }