hdu 3033 I love sneakers!（分组背包：每组至少一个）

I love sneakers!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 890    Accepted Submission(s): 381

Problem Description

After months of hard working, Iserlohn finally wins awesome amount of scholarship. As a great zealot of sneakers, he decides to spend all his money on them in a sneaker store.

There are several brands of sneakers that Iserlohn wants to collect, such as Air Jordan and Nike Pro. And each brand has released various products. For the reason that Iserlohn is definitely a sneaker-mania, he desires to buy at least one product for each brand.
Although the fixed price of each product has been labeled, Iserlohn sets values for each of them based on his own tendency. With handsome but limited money, he wants to maximize the total value of the shoes he is going to buy. Obviously, as a collector, he won’t buy the same product twice.
Now, Iserlohn needs you to help him find the best solution of his problem, which means to maximize the total value of the products he can buy.

 

Input

Input contains multiple test cases. Each test case begins with three integers 1<=N<=100 representing the total number of products, 1 <= M<= 10000 the money Iserlohn gets, and 1<=K<=10 representing the sneaker brands. The following N lines each represents a product with three positive integers 1<=a<=k, b and c, 0<=b,c<100000, meaning the brand’s number it belongs, the labeled price, and the value of this product. Process to End Of File.

 

Output

For each test case, print an integer which is the maximum total value of the sneakers that Iserlohn purchases. Print "Impossible" if Iserlohn's demands can’t be satisfied.

 

Sample Input

5 10000 3
1 4 6
2 5 7
3 4 99
1 55 77
2 44 66

 

Sample Output

255

 

         此题也是分组背包（每组至少放一个）！！！

      和普通的分组背包（每组至多放一个）不用，看了网上的解题报告，把后面两个循环调转一下就行，至于为什么，我也不知道怎么解释，直接套。

链接：http://acm.hdu.edu.cn/showproblem.php?pid=3033

代码：

#include <iostream>
#include <stdio.h>
#include <memory.h>
using namespace std;

const int INF = 99999999;

int max(int a, int b, int c)
{
    int M;
    M = a > b ? a : b;
    M = M > c ? M : c;
    return M;
}

int c[105], w[105], s[105];
int bag[15][10005];
int N, V, S;

void _gro_bag()
{
    int i, j, k;
    for(i = 1; i <= S; i++)
        for(j = 0; j <= V; j++)
            bag[i][j] = -INF;
    for(i = 1; i <= S; i++)
        for(k = 0; k < N; k++)
            for(j = V; j >= 0; j--)
                if(s[k] == i && j >= c[k])
                    bag[i][j] = max(bag[i][j], bag[i-1][j-c[k]]+w[k], bag[i][j-c[k]]+w[k]);
}

int main()
{
    int i;
    while(scanf("%d %d %d", &N, &V, &S) != EOF)
    {
        for(i = 0; i < N; i++)
            scanf("%d%d%d", &s[i], &c[i], &w[i]);
        _gro_bag();
        if(bag[S][V] < 0) printf("Impossible\n");
        else printf("%d\n", bag[S][V]);
    }

    return 0;
}