UVA 10130 SuperSale(dp + 01背包)

here 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

题意：有g个人去采购，每个人都有一定的负重量mw，有n样商品，每个商品都有价值p和重量w。要求所有人采购，

每个人每样商品只能采购一次，要求最后所有人采购的总价值最大。

思路：01背包问题。状态转移方程为dp[j] = max(dp[j - w[i]] + p[i], dp[j])。

代码：

#include <stdio.h>
#include <string.h>

int t, n, p[1005], w[1005], g, mw, i, j, dp[30005], sum;

int max(int a, int b) {
	return a > b ? a : b;
}
int main() {
	scanf("%d", &t);
	while (t --) {
		sum = 0;
		int s = 0;
		scanf("%d", &n);
		for (i = 0; i < n; i ++) {
			scanf("%d%d", &p[i], &w[i]);
			s += w[i];
		}
		memset(dp, 0, sizeof(dp));
		for (i = 0; i < n; i ++)
			for (j = s; j >= w[i]; j --) {
				if (dp[j - w[i]] || j - w[i] == 0)
					dp[j] = max(dp[j - w[i]] + p[i], dp[j]);
			}
			scanf("%d", &g);
			while (g --) {
				int Max = 0;
				scanf("%d", &mw);
				for (i = 0; i <= mw; i ++)
					Max = max(Max, dp[i]);				
				sum += Max;
			}
			printf("%d\n", sum);
	}
	return 0;
}

补充：软件开发 , C++ ,