hdu 5410 CRB and His Birthday（0-1背包+完全背包）

CRB and His Birthday

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1605    Accepted Submission(s): 755
[/b]

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

 

Author
KUT（DPRK）
解题思路：一个0-1背包和完全背包的融合，每种物品只有买第一种的时候会不同，之后完全一样，所以只需要在完全背包前面再加一个完全背包就好了~（^_^）

注意两种不同循环的方向~

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define maxn 1005
int t,n,m;
int w[maxn],a[maxn],b[maxn];
int dp[2005];
int main()
{
int i,j;
cin>>t;
while(t--)
{
cin>>m>>n;
for(i=0; i<n; i++)
cin>>w[i]>>a[i]>>b[i];
memset(dp,0,sizeof dp);
for(i=0; i<n; i++)
{
for(j=m; j>=w[i]; j--)//0-1
{
dp[j] = max(dp[j],dp[j-w[i]]+a[i]+b[i]);
}
for(j=w[i]; j<=m; j++)//完全背包
{
dp[j] = max(dp[j],dp[j-w[i]]+a[i]);
}
}
cout << dp[m] << endl;
}
//cout << "Hello world!" << endl;
return 0;
}