CRB and His Birthday(背包)

CRB and His Birthday

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

Total Submission(s): 357 Accepted Submission(s): 191

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：今天是CRB的生日，他的妈妈去商店给他买礼物，由于收银员是他妈妈的好朋友，所以收银员会按照不同礼 物的件数x赠与CRB的妈妈（a*x+b）块糖果。CRB的妈妈总共带了w元钱，总共有n种礼物。t组输入，每组先输入w 和n，接下来n行每行是该种礼物需要花费的价格和对应的a与b。求CRB的妈妈最多能获得多少糖果。

思路:因为每件物品都是无穷的所以是完全背包,但是只有在买第一件物品是多加b,所以将物品分成两部分,一部分是只有一件物品,进行01背包,一部分进行完全背包

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;

typedef unsigned long long LL;

int Dp[2100];

int n,m;

int main()
{
    int T;
    int w,a,b;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&m,&n);
        memset(Dp,0,sizeof(Dp));
        for(int i=1;i<=n;i++)
        {
            scanf("%d %d %d",&w,&a,&b);
            for(int j=m;j>=w;j--)//01背包
            {
                Dp[j]=max(Dp[j],Dp[j-w]+a+b);
            }
            for(int j=w;j<=m;j++)//完全背包
            {
                Dp[j]=max(Dp[j],Dp[j-w]+a);
            }
        }
        printf("%d\n",Dp[m]);
    }
    return 0;
}