poj 1170 Shopping Offers
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 4696 | Accepted: 1967 |
Description
In a shop each kind of product has a price. For example, the price of a flower is 2 ICU (Informatics Currency Units) and the price of a vase is 5 ICU. In order to attract more customers, the shop introduces some special offers.
A special offer consists of one or more product items for a reduced price. Examples: three flowers for 5 ICU instead of 6, or two vases together with one flower for 10 ICU instead of 12.
Write a program that calculates the price a customer has to pay for certain items, making optimal use of the special offers. That is, the price should be as low as possible. You are not allowed to add items, even if that would lower the price.
For the prices and offers given above, the (lowest) price for three flowers and two vases is 14 ICU: two vases and one flower for the reduced price of 10 ICU and two flowers for the regular price of 4 ICU.
Input
Output
Sample Input
2
7 3 2
8 2 5
2
1 7 3 5
2 7 1 8 2 10
Sample Output
14
Source
#include <stdio.h>
#include <string.h>
#pragma warning(disable:4996)
int b,c,k,p,s,n, a[], v[],w[];
int dp[+],id[],sum;
int main()
{
while (~scanf("%d", &b))
{
int six = ;
sum = ;
memset(id, -, sizeof id);
for (int i = ; i < b; i++)
{
scanf("%d%d%d", &c, &k, &p);
id[c] = i;
a[i] = k;
w[i] = six;
v[i] = p;
sum += w[i] * a[i];
six *= ;
}
for (int i = ; i <= sum; i++)dp[i] = << ;
scanf("%d", &s);
while (s--)
{
a[b] = <<;
w[b] = ;
scanf("%d", &n);
int s = ;
for (int i = ; i < n; i++)
{
scanf("%d%d", &c, &k);
if (id[c] == -)continue;
a[b] = a[b] < a[id[c]] / k ? a[b] : a[id[c]] / k;
w[b] += k*w[id[c]];
s += k*v[id[c]];
}
if (a[b] == <<)a[b] = ;
scanf("%d", &p);
v[b] = s<p?s:p;
b++;
}
for (int i = ; i < b; i++)
for (int k = ; k < a[i]; k++)
for (int j = sum; j >= w[i]; j--)
{
if (dp[j] > dp[j - w[i]] + v[i])dp[j] = dp[j - w[i]] + v[i];
}
printf("%d\n", dp[sum]);
//for (int i = 0; i <= sum; i++) printf("%d\n", dp[i]);
}
}
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