hdu 5410 CRB and His Birthday(01背包+完全背包)

CRB and His Birthday

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）

 

Source

2015 Multi-University Training Contest 10

参考博客：http://blog.csdn.net/keshuai19940722/article/details/47843927

思路：对于每一个礼物，第一次买是01背包，以后再买就是完全背包（因为要保持不论买多少件这个礼物，都只加一次bi）

代码：

#include<stdio.h>
#include<string.h>

#define maxn 2000+10
#define max(a,b) (a>b?a:b)
int dp[maxn];

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(dp,0,sizeof(dp));
int m,n;
scanf("%d%d",&m,&n);
int i,j,w,a,b;
for(i=0;i<n;i++)
{
scanf("%d%d%d",&w,&a,&b);
for(j=m;j>=w;j--)
dp[j]=max(dp[j],dp[j-w]+a+b);
for(j=w;j<=m;j++)
dp[j]=max(dp[j],dp[j-w]+a);
}
int ans=-1;
for(i=0;i<=m;i++)
ans=max(ans,dp[i]);
printf("%d\n",ans);
}
return 0;
}

总结：一开始以为最多买2000件，所以直接用多重背包写的，可是最后bi却无法保证加到正确的位置上0.0
原来这种包含只加一次和可以加多次的权值的话可以用01背包+完全背包的方法写