UVa 10130—SuperSale

最新推荐文章于 2020-03-02 22:37:01 发布

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本文深入探讨了在超市购物时如何合理分配家庭成员的携带能力，以实现最大价值的商品购买。通过实例分析，介绍了针对不同承载量的人群进行01背包问题的解决方法，从而帮助家庭在有限条件下最大化购物效益。

SuperSale

There is a SuperSale in a SuperHiperMarket. Every person can take only one object of each kind, i.e. one TV, one carrot, but for extra low price. We are going with a whole family to that SuperHiperMarket. Every person can take as many objects, as he/she can carry out from the SuperSale. We have given list of objects with prices and their weight. We also know, what is the maximum weight that every person can stand. What is the maximal value of objects we can buy at SuperSale?

Input Specification

The input consists of T test cases. The number of them (1<=T<=1000) is given on the first line of the input file.

Each test case begins with a line containing a single integer number N that indicates the number of objects (1 <= N <= 1000). Then follows N lines, each containing two integers: P and W. The first integer (1<=P<=100) corresponds to the price of object. The second integer (1<=W<=30) corresponds to the weight of object. Next line contains one integer (1<=G<=100) it’s the number of people in our group. Next G lines contains maximal weight (1<=MW<=30) that can stand this i-th person from our family (1<=i<=G).

Output Specification

For every test case your program has to determine one integer. Print out the maximal value of goods which we can buy with that family.

Sample Input

72 17

44 23

31 24

64 26

85 22

52 4

99 18

39 13

54 9

Output for the Sample Input

514

题意：m个人超市购物。没件物品限购一种，每个人有个承载量求m个人购得物品的最大价值

分析：对每个人的最大承载量进行0 1 背包

代码：

#include<stdio.h>
#include<string.h>
int p[1100],w[1100],dp[1100];
int n;
int ddp(int v)
{
    for(int i=0;i<n;i++)
        for(int j=v;j>=w[i];j--)
            if(dp[j]<dp[j-w[i]]+p[i])
                dp[j]=dp[j-w[i]]+p[i];
    return dp[v];
}
int main()
{

    int t,i,m,tem;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%d%d",&p[i],&w[i]);
        scanf("%d",&m);
        int ans=0;
        for(i=0;i<m;i++)
        {
            memset(dp,0,sizeof(dp));
            scanf("%d",&tem);
            ans+=ddp(tem);
        }
        printf("%d\n",ans);
    }
    return 0;
}