hdu5410(dp)

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

CRB and His Birthday

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2077    Accepted Submission(s): 978

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.

题意：有m元和n件物品，第i件物品的价格为w[i],价值为k*a[i]+b[i]（k为购买件数），求最大价值

题解：相当与一件物品，第一次购买的价值为a[i]+b[i]，之后他的价值就变为a[i]，

对于a[i]+b[i]的情况做一次01背包，对于a[i]的情况做一次完全背包dp[m]就是答案。

dp[i]：i元情况下所能购得的最大价值。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int t,m,n;int dp[2005];int w[2005],a[2005],b[2005];int main(){while(~scanf("%d",&t)){while(t--){memset(dp,0,sizeof(dp));scanf("%d%d",&m,&n);for(int i=1;i<=n;i++){scanf("%d%d%d",&w[i],&a[i],&b[i]);}//01背包for(int i=1;i<=n;i++){for(int j=m;j>=w[i];j--){dp[j]=max(dp[j],dp[j-w[i]]+a[i]+b[i]);}} //完全背包for(int i=1;i<=n;i++){for(int j=w[i];j<=m;j++){dp[j]=max(dp[j],dp[j-w[i]]+a[i]);}} printf("%d\n",dp[m]);}}return 0;}