先跑一遍01背包,再跑一遍多重背包

CRB and His Birthday

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

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

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月21日 星期五 13时29分32秒
File Name     :HDOJ5410.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int maxn=1100;

int m,n;
int w[maxn],A[maxn],B[maxn];
int dp[maxn*2];

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d%d",&m,&n);

		for(int i=1;i<=n;i++) scanf("%d%d%d",w+i,A+i,B+i);

		memset(dp,0,sizeof(dp));

		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]);
			}
		}

		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;
}

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