hdoj2191 珍惜现在，感恩生活（01背包 || 多重背包）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=2191

思路

由于每种大米可能不止一袋，所以是多重背包问题，可以直接使用解决多重背包问题的方法，也可以将多重背包转化为01背包后求解。

代码

01背包：

 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;

 const int N =  *  + ;
 int m[N], w[N];    //记录每袋大米的价格、重量
 int dp[N];

 int main()
 {
     //freopen("hdoj2191.txt", "r", stdin);
     int t;
     cin >> t;
     while (t--)
     {
         int n, k;
         cin >> n >> k;
         int cnt = ;    //共有cnt袋大米
         for (int i = ; i < k; i++)
         {
             int p, h, c;
             cin >> p >> h >> c;
             for (int j = ; j < c; j++)
             {
                 m[cnt] = p;
                 w[cnt] = h;
                 cnt++;
             }
         }

         memset(dp, , sizeof(dp));
         for (int i = ; i < cnt; i++)
         {
             for (int j = n; j >= m[i]; j--)
                 dp[j] = max(dp[j], dp[j - m[i]] + w[i]);
         }
         cout << dp[n] << endl;
     }
     return ;
 }

多重背包：

 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;

 const int N =  *  + ;
 int m[N], w[N], num[N];
 int dp[N];

 void zero_one_pack(int weight, int value, int capacity)
 {
     for (int i = capacity; i >= weight; i--)    //逆序
         dp[i] = max(dp[i], dp[i - weight] + value);
 }

 void complete_pack(int weight, int value, int capacity)
 {
     for (int i = weight; i <= capacity; i++)    //正序
         dp[i] = max(dp[i], dp[i - weight] + value);
 }

 void multiple_pack(int weight, int value, int amount, int capacity)
 {
     if (weight*amount >= capacity)
         complete_pack(weight, value, capacity);
     else
     {
         int k = ;
         while (k <= amount)
         {
             zero_one_pack(weight*k, value*k, capacity);
             amount -= k;
             k *= ;
         }
         zero_one_pack(weight*amount, value*amount, capacity);
     }

 }

 int main()
 {
     //freopen("hdoj2191.txt", "r", stdin);
     int t;
     cin >> t;
     while (t--)
     {
         int n, k;
         cin >> n >> k;
         for (int i = ; i < k; i++)
             cin >> m[i] >> w[i] >> num[i];

         memset(dp, , sizeof(dp));
         for (int i = ; i < n; i++)
             multiple_pack(m[i], w[i], num[i], n);
         cout << dp[n] << endl;
     }
     return ;
 }