HDU 5410 CRB and His Birthday

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5410

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

Hint

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

 

题目的大概意思是有M元钱，有N件物品，没件物品的花费是W[i]，第i件物品买X件能得到A[i]*X+B[i]块糖，没件物品买的数量没有上限

问这M元钱最多能得到多少块糖。

问题分析：没种物品买第一件能得到A[i]+B[i]块糖，以后每买一件只能多得到A[i]块糖，也就是每种物品买第一件比买后面的得到的糖多，

              根据贪心的思想很容易想到要先每种物品买一件再买大于一件，这样就把问题分成了两个阶段，第一阶段每种物品最多只买一件，

              0/1背包，第二阶段，没种物品能买无限件，完全背包。

#include <iostream>
#include <cstring>
using namespace std;

int ma(int a,int b)
{
    if (a>b)
    return a;
    else
    return b;
}
int main()
{
    int ans[12000],v,a[1100],b[1100],c[1100],n,m,T;
    cin>>T;
    while (T--)
    {
          cin>>v>>m;
          for (int i=1;i<=m;i++)
          cin>>c[i]>>a[i]>>b[i];
          memset(ans,0,sizeof(ans));
          for (int i=1;i<=m;i++)//处理第一阶段，0/1背包 
          {
              for (int j=v;j>=c[i];j--)
              {
                  ans[j]=ma(ans[j],ans[j-c[i]]+a[i]+b[i]);              
              }
          }
          for (int i=1;i<=m;i++)//处理第二阶段，完全背包。 
          {
              for (int j=c[i];j<=v;j++)
              {
                   ans[j]=ma(ans[j],ans[j-c[i]]+a[i]);
              }
          }
          cout <<ans[v]<<endl;
    }
    return 0;
}