HDU 5410 CRB and His Birthday（DP）

CRB and His Birthday

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 83    Accepted Submission(s): 45

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.
   
    
  

 

Author

KUT（DPRK）

 

Source

2015 Multi-University Training Contest 10

点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

const int N = 2010;

int dp[N];
int p[N],w[N],k[N];
int n,m;
int flag[N][N];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&p[i],&w[i],&k[i]);
        }
        int maxx = 0;
        memset(flag,0,sizeof(flag));
        memset(dp,0,sizeof(dp));
        for(int i=0; i<m; i++)
        {
            for(int j=n; j>=p[i]; j--)
            {
                dp[j] = max(dp[j],dp[j-p[i]]+w[i]+k[i]);
                maxx = max(dp[j],maxx);
            }
        }
        for(int i=0;i<m;i++){
            for(int j=p[i];j<=n;j++){
                dp[j] = max(dp[j],dp[j-p[i]]+w[i]);
                maxx = max(dp[j],maxx);
            }
        }
        printf("%d\n",maxx);
    }

    return 0;
}