CRB and His Birthday(hdu 5410)

CRB and His Birthday

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1112    Accepted Submission(s): 559

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

题意：

买东西，给你N的钱M种类的东西

每类东西，如果买就送A*x+B的蜡烛。

x为购买数，问最多可以获得多少蜡烛。

思路：因为送A*x+B的蜡烛，如果没有后面的常数B,就相当于价值为A,所以直接完全背包就可以解决。

但是后面多了个常数B,所以还有考虑买不买的情况，所以01背包先跑一遍，再跑一遍完全背包就可以了。

O(M*N);

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<math.h>
 6 #include<queue>
 7 #include<string.h>
 8 #include<stack>
 9 #include<vector>
10 #include<map>
11 #define sc(x) scanf("%I64d",&x)
12 #define pr(x) printf("%I64d",x)
13 #define prr(x) printf("%I64d\n",x)
14 #define prrr(x) printf(" %I64d",x)
15 #define FOR(i,p,q) for(int i=p;i<=q;i++)
16 typedef struct pp
17 {
18     int x;
19     int y;
20     int z;
21 }ss;
22 ss aa[3000];
23 int dp[3000];
24 using namespace std;
25 int main(void)
26 {
27     int n,i,j,k,p,q;
28     int N,M;
29     scanf("%d",&k);
30     while(k--)
31     {
32         scanf("%d %d",&N,&M);
33         memset(dp,0,sizeof(dp));
34         for(i=1;i<=M;i++)
35         {
36             scanf("%d %d %d",&aa[i].x,&aa[i].y,&aa[i].z);
37         }
38         for(i=1;i<=M;i++)
39         {
40             for(j=N;j>=aa[i].x;j--)
41             {
42                 dp[j]=max(dp[j],dp[j-aa[i].x]+aa[i].z+aa[i].y);
43             }
44         }
45         for(i=1;i<=M;i++)
46         {
47             for(j=aa[i].x;j<=N;j++)
48             {
49                 dp[j]=max(dp[j],dp[j-aa[i].x]+aa[i].y);
50             }
51         }
52         printf("%d\n",dp[N]);
53     }
54     return 0;
55 }