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 of the kind, i.e. one TV, one carrot, but for extra. We are going with a whole family to that superhipermarket. Every 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 this 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 "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 "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
This article URL address: http://www.bianceng.cn/Programming/sjjg/201410/45377.htm
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
Water and water.
Complete code:
/*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;
}