01背包+完全背包 HDU - 5410 

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#01背包+完全背包 HDU - 5410

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本文介绍了一个关于如何在有限预算内购买礼物以获得最大数量糖果的问题。通过使用01背包和完全背包算法，解决了选择购买何种礼物组合以最大化收益的问题。

Today is CRB's birthday. His mom decided to buy many presents for her lovely son.
She went to the nearest shop with M
M
Won(currency unit).
At the shop, there are N
N
kinds of presents.
It costs Wi
Wi
Won to buy one present of i
i
-th kind. (So it costs k
k
× Wi
Wi
Won to buy k
k
of them.)
But as the counter of the shop is her friend, the counter will give Ai × x + Bi
Ai × x + Bi
candies if she buys x
x
(x
x
>0) presents of i
i
-th kind.
She wants to receive maximum candies. Your task is to help her.
1 ≤ T
T
≤ 20
1 ≤ M
M
≤ 2000
1 ≤ N
N
≤ 1000
0 ≤ Ai, Bi
Ai, Bi
≤ 2000
1 ≤ Wi
Wi
≤ 2000

Input

There are multiple test cases. The first line of input contains an integer T
T
, indicating the number of test cases. For each test case:
The first line contains two integers M
M
and N
N
.
Then N
N
lines follow, i
i
-th line contains three space separated integers Wi
Wi
, Ai
Ai
and Bi
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.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
#define LL long long
#define MAX 2005
using namespace std;
int dp[MAX];
int w[MAX],a[MAX],b[MAX];
int t,m,n;
int main()
{
    scanf("%d",&t);
    while(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]);   //先通过01背包来确定拿不拿这件商品的最大价值
        }
        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]);  //用完全背包计算拿了x件商品的最大价值（a[i]*x+b[i])
        }
        printf("%d\n",dp[m]);
    }


    return 0;
}

/*题意：
有M多的钱 去买n种商品
每种商品的价格为w  如果你买x个这种商品 你会得到 a*x+b个糖果
问最多能得到多少糖果*/

/*思路：
对于每件商品 你如果买一件 那么你会得到a+b个糖果
如果再次基础上每一次的购买你就只能得到a个糖果
所以可以看所这种商品有两种价值
第一种价值只有一次 用01背包解决
第二种价值可以无限次购买 用完全背包解决 */