UVa 10130 SuperSale(DP_01背包)

最新推荐文章于 2020-03-02 22:37:01 发布

原创最新推荐文章于 2020-03-02 22:37:01 发布 · 774 阅读

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本文介绍了一个关于SuperSale购物的问题：一家人在超市面对各种价格和重量的商品，在每个人的携带能力限制下，如何选择商品以达到购买价值最大化。通过01背包问题的解决思路，使用动态规划算法确定每个人能携带物品的最大价值。

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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

每个人每个商品只能拿一件要拿价值尽量高的商品很简单的01背包题目求出每个人可以拿的最大值加起来就是结果了

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 1005
int p[maxn],w[maxn],d[maxn],pe,t,n,g;
int main()
{
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;++i)
            scanf("%d%d",&p[i],&w[i]);

        scanf("%d",&g);
        int ans=0;
        for(int k=1;k<=g;++k)
            {
                memset(d,0,sizeof(d));
                scanf("%d",&pe);
                for(int i=1;i<=n;++i)
                {
                    for(int j=pe;j>=w[i];j--)
                        d[j]=max(d[j],d[j-w[i]]+p[i]);
                }
                ans+=d[pe];
            }
        printf("%d\n",ans);
    }
    return 0;
}