UVa 10130 SuperSale (0-1背包)

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本文介绍了一个关于家庭成员如何在限定重量下最大化购物价值的问题。通过动态规划算法，解决每个家庭成员如何选择商品以达到总价值最大化的挑战。输入包括商品的价格、重量及家庭成员的承重能力。

10130 - SuperSale

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1071

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 Nlines, 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

水水。

完整代码：

/*0.072s*/

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int p[1005], w[1005], dp[35];

int main()
{
	int t, n, i, j, g, sum, maxw;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		for (i = 0; i < n; ++i)
			scanf("%d%d", &p[i], &w[i]);
		scanf("%d", &g);
		sum = 0;
		while (g--)
		{
			memset(dp, 0, sizeof(dp));
			scanf("%d", &maxw);
			for (i = 0; i < n; ++i)
				for (j = maxw; j >= w[i]; --j)
					dp[j] = max(dp[j], dp[j - w[i]] + p[i]);
			sum += dp[maxw];
		}
		printf("%d\n", sum);
	}
	return 0;
}