HDU 5410 CRB and His Birthday(完全背包)——多校练习10

完全背包问题与糖果获取最大化的解题策略

最新推荐文章于 2021-07-27 09:06:55 发布

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#acm #动态规划 #完全背包

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本文详细解析了如何运用完全背包的概念来解决CRB生日时妈妈通过购买礼物获取最大糖果数量的问题。文章包括解题思路、代码实现及几个样例输入输出，帮助读者理解并掌握该类问题的解决方法。

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

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：CRB生日，妈妈要给他买礼物，妈妈有M元钱，这家店有N种礼物，因为店长和妈妈是熟人，所以若第i种礼物买x件的话，店长会给妈妈Ai*x+Bi颗糖果，现给出每种礼物的单价、Ai值与Bi值，问妈妈最多能拿到多少颗糖果。

放上出题人的解题报告

解题思路：刚拿到这题的时候就知道是完全背包的问题，因为每件礼物的数量无限多，我可以不取、取一件或取多件；而01背包的话是每种商品只有取或不取，没有多件之说；至于多重背包，则是每件商品的件数是有限的，告诉你的。关于完全背包，不懂的可以点链接，表示这个人讲解得挺好的。

该题知道是完全背包之后，问题还并没有解决，因为单种礼物能获得的糖果数与购买的礼物件数成线性关系，所以一件与多件的差距在于a[i]的倍数，而买或不买的差距则在于a[i]+b[i]。因此，b[i]的贡献仅在于买该种礼物的第一件

提供几组样例参考

Input

100
100 2
10 2 1
10 1 2

Output
22

Input
100
50 3
10 0 1
15 0 2
11 0 2

Output
5

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 1005;
const int inf = 1000000000;
const int mod = 1000000007;
int s[N*2][2],w[N],a[N],b[N];
int main()
{
    int t,m,n,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&m,&n);
        memset(s,0,sizeof(s));
        for(i=1; i<=n; i++)
            scanf("%d%d%d",&w[i],&a[i],&b[i]);
        for(i=1; i<=n; i++)
            for(j=0; j<=m; j++)
            {
                s[j][0]=max(s[j][0],s[j][1]);
                if(j>=w[i])
                    s[j][1]=max(s[j-w[i]][0]+a[i]+b[i],s[j-w[i]][1]+a[i]);
                else
                    s[j][1]=0;
            }
        printf("%d\n",max(s[m][0],s[m][1]));
    }
    return 0;
}

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