HDU 5410 CRB and His Birthday(完全背包，01背包)

CRB and His Birthday

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2380    Accepted Submission(s): 1120

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

1100 210 2 120 1 1

 

Sample Output

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

 

Author

KUT（DPRK）

 

Source

2015 Multi-University Training Contest 10

思路:f(x)=ax+b=a(x-1)+a+b，虽然买a物品必须要先买a+b好像有些问题，但因为a+b>a且两者价值相等，所以买a较优时必定会存在买a+b的情况，所以这么拆分没什么影响。

代码:

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;#define maxn 2005int w[maxn],a[maxn],b[maxn],dp[maxn];int main(){    int C,M,n;    scanf("%d",&C);    while(C--)    {        scanf("%d%d",&M,&n);        int i,j;        for(i=0;i<n;i++)            scanf("%d%d%d",&w[i],&a[i],&b[i]);        for(i=0;i<n;i++)        {            for(j=M;j>=w[i];j--)            {                dp[j]=max(a[i]+b[i]+dp[j-w[i]],dp[j]);            }            for(j=w[i];j<=M;j++)            {                dp[j]=max(a[i]+dp[j-w[i]],dp[j]);            }        }        printf("%d\n",dp[M]);        memset(dp,0,sizeof(dp));    }    return 0;}