[动态规划]UVA10130 - SuperSale

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

2

3

72 17

44 23

31 24

1

26

6

64 26

85 22

52 4

99 18

39 13

54 9

4

23

20

20

26

Output for the Sample Input

72

514

题意：在一个商场有个超级拍卖。每个人每样商品只能拿一件。现在有一家人到超级拍卖会去拿商品，只要他搬的懂。我们手上有所有商品的价格以及重量，我们也知道这家人每个人最多能搬的重量，请问这家人能从超级拍卖会搬到的最大价格是多少？

思路：典型的0/1背包问题，只不过是换成了N个人。

#include<iostream>
#include<cstring>

using namespace std;

int cost[1010],val[1010],per[1010],dp[1010][50];

int main()
    {
        int num;
        cin>>num;
        while(num--)
            {
                int n,m;
                cin>>n;
                for(int i=0;i<n;i++)
                    cin>>val[i]>>cost[i];
                cin>>m;
                for(int i=0;i<m;i++)
                    cin>>per[i];
                int sum=0;
                for(int i=0;i<m;i++)
                    {
                        memset(dp,0,sizeof(dp));
                        for(int j=0;j<n;j++)
                            {
                                for(int k=0;k<=per[i];k++)
                                    {
                                        if(k<cost[j]) dp[j+1][k]=dp[j][k];
                                        else dp[j+1][k]=max(dp[j][k],dp[j][k-cost[j]]+val[j]);
                                    }
                            }
                        sum=sum+dp[n][per[i]];
                    }
                cout<<sum<<endl;
            }
        return 0;
    }

[动态规划]UVA10130 - SuperSale,布布扣,bubuko.com

[动态规划]UVA10130 - SuperSale

原文：http://blog.csdn.net/zju_ziqin/article/details/24183413