poj2392

多重背包，多重背包可以转化为完全背包和01背包。如果某东西的总体积大于包体积，则可以当成是完全背包。否则按物品体积的1，2，4...倍分别进行01背包。这样就相当于构成了所有的可能情况，例如5 ＝ 4 ＋ 1， 7 ＝ 1 ＋ 2 ＋ 4。都可以由这些数字构成。

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

#define maxn 405
#define maxm 40005

struct Block
{
    int count, h, limit;
}block[maxn];

int n;
bool f[maxm];

bool operator < (const Block &a, const Block &b)
{
    return a.limit < b.limit;
}

void compackage(int h, int limit)
{
    for (int i = h; i <=limit; i++)
        f[i] = f[i] || f[i - h];
}

void zerpackage(int h, int limit)
{
    for (int i = limit; i >=h; i--)
        f[i] = f[i] || f[i - h];
}

void mulpackage(int count, int h, int limit)
{
    if (h * count >= limit)
    {
        compackage(h, limit);
        return;
    }
    int i = 1;
    while (count >= i)
    {
        zerpackage(h * i, limit);
        count -= i;
        i <<= 1;
    }
    zerpackage(h * count, limit);
}

int main()
{
    //freopen("t.txt", "r", stdin);
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d%d%d", &block[i].h, &block[i].limit, &block[i].count);
    sort(block, block + n);
    memset(f, 0, sizeof(f));
    f[0] = true;
    for (int i = 0; i < n; i++)
        mulpackage(block[i].count, block[i].h, block[i].limit);
    for (int i = block[n - 1].limit; i >= 0; i--)
        if (f[i])
        {
            printf("%d\n", i);
            break;
        }
    return 0;
}