多重背包（二进制模板+普通解法）

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

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <cstdio>
 5
 6 using namespace std;
 7 const int MAX = 100 + 10;
 8 int V;
 9 int f[MAX],Price[MAX],Weight[MAX],Amount[MAX];
10 void OneZeorPage(int cost, int weight)
11 {
12     for(int i = V; i >= cost; i--)
13         f[i] = max(f[i], f[i - cost] + weight);
14 }
15 void CompletePage(int cost, int weight)
16 {
17     for(int i = cost; i <= V; i++)
18         f[i] = max(f[i], f[i - cost] + weight);
19
20 }
21 void multiplePage(int cost,int weight,int amount)
22 {
23     if(amount * cost >= V)
24     {
25         CompletePage(cost,weight);
26         return;
27     }
28     int k = 1;
29     while(k < amount)
30     {
31         OneZeorPage(k * cost, k * weight);
32         amount -= k;
33         k <<= 1;
34     }
35     if(amount > 0)
36         OneZeorPage(amount * cost, amount * weight);
37     return;
38 }
39 int main()
40 {
41     int n,m,t;
42     scanf("%d", &t);
43     while(t--)
44     {
45         scanf("%d%d", &n,&m);
46         for(int i = 1; i <= m; i++)
47             scanf("%d%d%d", &Price[i],&Weight[i],&Amount[i]);
48         memset(f,0,sizeof(f));
49         V = n;
50         for(int i = 1; i <= m; i++)
51             multiplePage(Price[i],Weight[i],Amount[i]);
52         printf("%d\n",f[n]);
53     }
54     return 0;
55 }

二进制优化

那为什么会有完全背包和01 背包的不同使用加判断呢？原因也很简单啊，当数据很大，大于背包的容纳量时，我们就是在这个物品中取上几件就是了，取得量时不知道的，也就理解为无限的啦，这就是完全背包啦，反而小于容纳量的就是转化为01背包来处理就是了，可以大量的省时间。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <cstdio>
 5
 6 using namespace std;
 7 const int MAX = 100 + 10;
 8 int V;
 9 int f[MAX],Price[MAX],Weight[MAX],Amount[MAX];
10 int main()
11 {
12     int n,m,t;
13     scanf("%d", &t);
14     while(t--)
15     {
16         scanf("%d%d", &n,&m);
17         for(int i = 1; i <= m; i++)
18             scanf("%d%d%d", &Price[i],&Weight[i],&Amount[i]);
19         memset(f,0,sizeof(f));
20         V = n;
21         for(int i = 1; i <= m; i++)
22         {
23             for(int k = 1; k <= Amount[i]; k++)
24             {
25                 for(int j = V; j >= Price[i]; j--)
26                 {
27                     f[j] = max(f[j], f[j - Price[i]] + Weight[i]);
28                 }
29             }
30         }
31         printf("%d\n",f[n]);
32     }
33     return 0;
34 }

朴素的算法