hdu 5410 CRB and His Birthday(01+多重揹包 dp)

CRB and His Birthday

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1188    Accepted Submission(s): 591

Problem Description

Today is CRB's birthday. His mom decided to buy many presents for her lovely son.
She went to the nearest shop with M Won(currency unit).
At the shop, there are N kinds of presents.
It costs Wi Won to buy one present of i-th kind. (So it costs k × Wi Won to buy k of them.)
But as the counter of the shop is her friend, the counter will give Ai × x + Bi candies if she buys x(x>0) presents of i-th kind.
She wants to receive maximum candies. Your task is to help her.
1 ≤ T ≤ 20
1 ≤ M ≤ 2000
1 ≤ N ≤ 1000
0 ≤ Ai, Bi ≤ 2000
1 ≤ Wi ≤ 2000

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains two integers M and N.
Then N lines follow, i-th line contains three space separated integers Wi, Ai and Bi.

 

Output

For each test case, output the maximum candies she can gain.

 

Sample Input

1 100 2 10 2 1 20 1 1

 

Sample Output

21

solution：

这道题我们可以先跑一遍01揹包，在跑一遍多重揹包。

思路：非常好的一道01揹包和完全揹包结合的题目 首先，对于第i件商品，如果只买1个，得到的价值是Ai+Bi 如果在买1个的基础上再买，得到的价值就是Ai 也就是说，除了第一次是Ai+Bi，以后购买都是Ai 那么，我们能否将i商品拆分成两种商品，其中两种商品的代价都是Wi, 第一种的价值是Ai+Bi，但是只允许买一次 第二种的价值是Ai，可以无限次购买 接下来我们来讨论这样拆的正确性 理论上来讲，买第二种之前，必须要买第一种 但是对于这道题，由于Ai+Bi>=Ai是必然的，因为Bi肯定是非负 所以对于代价相同，价值大的肯定会被先考虑 换句话来说，如果已经开始考虑第二种商品了，那么第一种商品就肯定已经被添加到揹包里了

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 2500;
int dp[maxn], a[maxn], b[maxn],w[maxn];
int n, m;
int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        memset(dp, 0, sizeof(dp));
        scanf("%d%d", &m,&n);
        for (int i = 1; i <= n; i++)
            scanf("%d%d%d",&w[i], &a[i], &b[i]);
        for (int i = 1; i <= n; i++)
        {
            for (int j = m; j >= w[i]; j--)
                dp[j] = max(dp[j], dp[j - w[i]] + a[i] + b[i]);
            for (int j = w[i]; j <= m; j++)
                dp[j] = max(dp[j], dp[j - w[i]] + a[i]);
        }
        printf("%d\n", dp[m]);
    }
    return 0;
}