hdu5410~CRB and His Birthday(01背包+完全背包)

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CRB and His Birthday

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1874    Accepted Submission(s): 884


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 WiAi 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
 

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#include <iostream>#include <iomanip>#include<stdio.h>#include<string.h>#include<stack>#include<stdlib.h>#include<queue>#include<math.h>#include<algorithm>#include<vector>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;int w[2005],a[2005],b[2005],sum[2005];int main(){   int s;   scanf("%d",&s);   while(s--)   {       int m,n;       scanf("%d%d",&m,&n);       for(int i=1;i<=n;i++)       {           scanf("%d%d%d",&w[i],&a[i],&b[i]);       }       mem(sum,0);       for(int i=1;i<=n;i++)       {           for(int j=m;j>=w[i];j--)           {               sum[j]=max(sum[j],sum[j-w[i]]+a[i]+b[i]);           }           for(int j=w[i];j<=m;j++)           {               sum[j]=max(sum[j],sum[j-w[i]]+a[i]);           }       }       printf("%d\n",sum[m]);   }}



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