01背包+完全背包 HDU - 5410 

Today is CRB's birthday. His mom decided to buy many presents for her lovely son.

She went to the nearest shop with M

M

Won(currency unit).

At the shop, there are N

N

kinds of presents.

It costs Wi

Wi

Won to buy one present of i

i

-th kind. (So it costs k

k

× Wi

Wi

Won to buy k

k

of them.)

But as the counter of the shop is her friend, the counter will give Ai × x + Bi

Ai × x + Bi

candies if she buys x

x

(x

x

>0) presents of i

i

-th kind.

She wants to receive maximum candies. Your task is to help her.

1 ≤ T

T

≤ 20

1 ≤ M

M

≤ 2000

1 ≤ N

N

≤ 1000

0 ≤ Ai, Bi

Ai, Bi

≤ 2000

1 ≤ Wi

Wi

≤ 2000

Input

There are multiple test cases. The first line of input contains an integer T

T

, indicating the number of test cases. For each test case:

The first line contains two integers M

M

and N

N

.

Then N

N

lines follow, i

i

-th line contains three space separated integers Wi

Wi

, Ai

Ai

and Bi

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

Hint

CRB's mom buys 10 presents of first kind, and receives 2 × 10 + 1 = 21 candies.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
#define LL long long
#define MAX 2005
using namespace std;
int dp[MAX];
int w[MAX],a[MAX],b[MAX];
int t,m,n;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&m,&n);
for(int i=1;i<=n;i++)
scanf("%d%d%d",&w[i],&a[i],&b[i]);
memset(dp,0,sizeof(dp));
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]); //先通过01背包来确定拿不拿这件商品的最大价值
}
for(int i=1;i<=n;i++)
for(int j=w[i];j<=m;j++)
{
dp[j]=max(dp[j],dp[j-w[i]]+a[i]); //用完全背包计算拿了x件商品的最大价值（a[i]*x+b[i])
}
printf("%d\n",dp[m]);
}

return 0;
}

/*题意：
有M多的钱 去买n种商品
每种商品的价格为w 如果你买x个这种商品 你会得到 a*x+b个糖果
问最多能得到多少糖果*/

/*思路：
对于每件商品 你如果买一件 那么你会得到a+b个糖果
如果再次基础上每一次的购买你就只能得到a个糖果
所以可以看所这种商品有两种价值
第一种价值只有一次 用01背包解决
第二种价值可以无限次购买 用完全背包解决 */