Day 3 DP

1. P1015 精卫填海

01背包 + 判断

 #include <iostream>
 using namespace std;
 const int MAXN = , MAXV = ;
 int v, n, m, f[MAXV], w[MAXN], c[MAXN];//f[i]为消耗体力为i时最大的石头体积
 int main()
 {
     cin >> v >> n >> m;
     for(int i = ; i <= n; ++i)
         cin >> c[i] >> w[i];
     for(int i = ; i <= n; ++i)
         for(int j = m; j >= w[i]; --j)
             f[j] = max(f[j], f[j - w[i]] + c[i]);
     for(int i = ; i <= m; ++i)
         if(f[i] >= v)
         {
             cout << m - i << endl;
             return ;
         }
     cout << "Impossible" << endl;
     return ;
 } 

2. P1507 NASA的食物

类似于01背包，但相比01背包的唯一的体积(或质量)的限制条件增加了质量(或体积)

 #include <iostream>
 using namespace std;
 const int MAXN = , MAXM = , MAXK = ;
 int m, n, k, f[MAXN][MAXM], a[MAXK], b[MAXK], c[MAXK];
 int main()
 {
     cin >> m >> n >> k;
     for(int i = ; i <= k; ++i)
         cin >> a[i] >> b[i] >> c[i];
     for(int i = ; i <= k; ++i)
         for(int u = m; u >= a[i]; --u)
             for(int v = n; v >= b[i]; --v)
                 f[u][v] = max(f[u][v], f[u - a[i]][v - b[i]] + c[i]);
     cout << f[m][n] << endl;
     return ;
 }

By ZRQ