HDOJ 3033

I love sneakers!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3096    Accepted Submission(s): 1269

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
   
   
   
   

 

Source

2009 Multi-University Training Contest 13 - Host by HIT

 

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gaojie

 

一道不错的分组背包。借用一下某大神的讲解吧

题意：

某人要买鞋，有k个品牌，每个品牌有j个款，每款都有标价和价值，要求已经M元内，每个品牌至少买一双鞋的最大价值和。

思路：

1. 要求每一组之中至少有一个被选中，和之前的最多有一个稍有区别，需要把题目再次细分。

2. dp[i][j] 表示选定 i 个品牌并且花费为 j 的最大价值，dp[i][j] 为正数表示状态存在，为负数表示状态不存在。

3. dp[i][j] = max(dp[i][j], dp[i][j - v_k] + w_k);      第 i 类品牌有选择并且要选择第 k 件物品。（不选择第 k 件物品是相等的，所以略过转移方程）

   dp[i][j] = max(dp[i][j], dp[i - 1][j - v_k] + w_k);  第 i 类品牌前面没有选择并且要选择第 k 件物品。

4. 由于状态是从 0 到 i 的且 dp[0][j] = 0，其他为 -INFS 。所以只有当第一类品牌的状态存在时，才能推导出来第二类品牌的存在状态，以此类推。

5. 题目中有 2 个陷阱，一是可能会存在某品牌数量为 0 的情况，另外会存在费用或价格为 0 的情况，所以状态转移方程不能调换。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <iomanip>

using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1|1
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
#define mid ((l + r) >> 1)
#define mk make_pair
const int MAXN = 10000 + 50;
const int maxw = 100 + 20;
const int MAXNNODE = 1000 +10;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
#define mod 10007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pii;
const D e = 2.718281828459;
int dp[12][MAXN] , n , m , k;
int  cost , value , id;
vector <pii> brand[12];
int main()
{
    //ios::sync_with_stdio(false);
#ifdef Online_Judge
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif // Online_Judge
    while(~scanf("%d%d%d" , &n , &m , &k))
    {
        FORR(i , 1 , k)brand[i].clear();
        FOR(i ,0 , n)
        {
            scanf("%d%d%d" , &id , &cost , &value);
            brand[id].push_back(make_pair(cost , value));
        }
        clr(dp[0] , 0);
        int flag = 0;
        FORR(i , 1 , k)
        {
            FORR(j ,0 , m)dp[i][j] = -INF;
            if(!brand[i].empty())flag++;
            FOR(j , 0 ,brand[i].size())
            {
                cost = brand[i][j].first;
                value = brand[i][j].second;
                REPP(v , m , cost)
                {
                    dp[i][v] = max(dp[i][v] , dp[i][v - cost] + value);
                    dp[i][v] = max(dp[i][v] , dp[i - 1][v - cost] + value);
                    
                }
            }
        }
        if(dp[k][m] > 0&& flag == k) printf("%d\n" , dp[k][m]);
        else printf("Impossible\n");
    }
    return 0;
}