多重背包（二进制模板+普通解法）

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

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;
const int MAX = 100 + 10;
int V;
int f[MAX],Price[MAX],Weight[MAX],Amount[MAX];
void OneZeorPage(int cost, int weight)
{
    for(int i = V; i >= cost; i--)
        f[i] = max(f[i], f[i - cost] + weight);
}
void CompletePage(int cost, int weight)
{
    for(int i = cost; i <= V; i++)
        f[i] = max(f[i], f[i - cost] + weight);

}
void multiplePage(int cost,int weight,int amount)
{
    if(amount * cost >= V)
    {
        CompletePage(cost,weight);
        return;
    }
    int k = 1;
    while(k < amount)
    {
        OneZeorPage(k * cost, k * weight);
        amount -= k;
        k <<= 1;
    }
    if(amount > 0)
        OneZeorPage(amount * cost, amount * weight);
    return;
}
int main()
{
    int n,m,t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n,&m);
        for(int i = 1; i <= m; i++)
            scanf("%d%d%d", &Price[i],&Weight[i],&Amount[i]);
        memset(f,0,sizeof(f));
        V = n;
        for(int i = 1; i <= m; i++)
            multiplePage(Price[i],Weight[i],Amount[i]);
        printf("%d
",f[n]);
    }
    return 0;
}

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。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;
const int MAX = 100 + 10;
int V;
int f[MAX],Price[MAX],Weight[MAX],Amount[MAX];
int main()
{
    int n,m,t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n,&m);
        for(int i = 1; i <= m; i++)
            scanf("%d%d%d", &Price[i],&Weight[i],&Amount[i]);
        memset(f,0,sizeof(f));
        V = n;
        for(int i = 1; i <= m; i++)
        {
            for(int k = 1; k <= Amount[i]; k++)
            {
                for(int j = V; j >= Price[i]; j--)
                {
                    f[j] = max(f[j], f[j - Price[i]] + Weight[i]);
                }
            }
        }
        printf("%d
",f[n]);
    }
    return 0;
}