01背包+完全背包 HDU

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 2005using 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 */