HDU 3033 I love sneakers! (分组背包 每组至少选一个)

I love sneakers!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5124    Accepted Submission(s): 2131

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 31 4 62 5 73 4 991 55 772 44 66

 

Sample Output

255

 

Source

2009 Multi-University Training Contest 13 - Host by HIT

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

题目大意：n双鞋子属于k个品牌，每双鞋子有一个价格和价值，问在钱数为m的情况下每个牌子至少买一个，每双至多买一个时的最大总价值

题目分析：分组背包，设dp[i][j]为用了j钱买了前i个品牌时的最大价值，转移时分两种情况
1.当前品牌不是第一次买     =>   dp[i][j] = max(dp[i][j], dp[i][j - w[i]] + val[i])
2.当前品牌是第一次买         =>  dp[i][j] = max(dp[i][j], dp[i - 1][j - w[i]] + val[i])
上面的顺序不能颠倒，因为代价若为0，则会出现同一双买两次的情况
初始时dp[0][j]为0，其余均为-1，且每次转移时必须从非-1态转移

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int const MAX = 105;int n, m, K;int dp[MAX][10005];struct DATA {    int b, p, val;}d[MAX];int main() {    while(scanf("%d %d %d", &n, &m, &K) != EOF) {        for (int i = 0; i < n; i ++) {            scanf("%d %d %d", &d[i].b, &d[i].p, &d[i].val);        }        memset(dp, -1, sizeof(dp));        for (int j = 0; j <= m; j ++) {            dp[0][j] = 0;        }        for (int k = 1; k <= K; k ++) {            for (int i = 0; i < n; i ++) {                if (d[i].b == k) {                    for (int j = m; j >= d[i].p; j --) {                        if (dp[k][j - d[i].p] != -1) {                            dp[k][j] = max(dp[k][j], dp[k][j - d[i].p] + d[i].val);                        }                        if (dp[k - 1][j - d[i].p] != -1) {                            dp[k][j] = max(dp[k][j], dp[k - 1][j - d[i].p] + d[i].val);                        }                    }                }            }        }        if (dp[K][m] == -1) {            printf("Impossible\n");        }        else {            printf("%d\n", dp[K][m]);        }    }}