题目地址：http://poj.org/problem?id=1170
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
Your program is to read from standard input. The first line contains the number b of different kinds of products in the basket (0 <= b <= 5). Each of the next b lines contains three values c, k, and p. The value c is the (unique)
product code (1 <= c <= 999). The value k indicates how many items of this product are in the basket (1 <= k <= 5). The value p is the regular price per item (1 <= p <= 999). Notice that all together at most 5*5=25 items can be in the basket. The b+2nd line
contains the number s of special offers (0 <= s <= 99). Each of the next s lines describes one offer by giving its structure and its reduced price. The first number n on such a line is the number of different kinds of products that are part of the offer (1
<= n <= 5). The next n pairs of numbers (c,k) indicate that k items (1 <= k <= 5) with product code c (1 <= c <= 999) are involved in the offer. The last number p on the line stands for the reduced price (1 <= p <= 9999). The reduced price of an offer is less
than the sum of the regular prices.
product code (1 <= c <= 999). The value k indicates how many items of this product are in the basket (1 <= k <= 5). The value p is the regular price per item (1 <= p <= 999). Notice that all together at most 5*5=25 items can be in the basket. The b+2nd line
contains the number s of special offers (0 <= s <= 99). Each of the next s lines describes one offer by giving its structure and its reduced price. The first number n on such a line is the number of different kinds of products that are part of the offer (1
<= n <= 5). The next n pairs of numbers (c,k) indicate that k items (1 <= k <= 5) with product code c (1 <= c <= 999) are involved in the offer. The last number p on the line stands for the reduced price (1 <= p <= 9999). The reduced price of an offer is less
than the sum of the regular prices.
Output
Your program is to write to standard output. Output one line with the lowest possible price to be paid.
Sample Input
2
7 3 2
8 2 5
2
1 7 3 5
2 7 1 8 2 10
Sample Output
14
谨以此来纪念蛋疼六重循环！！！
#include <stdio.h>
#include <string.h>
#include <limits.h> int code[6]; //商品代码
int num[6]; //商品数量
int price[6]; //商品价格
int special_num[100][6]; //促销项目各个商品数量
int special_cnt[100]; //促销项目的商品数量
int special_price[100]; //促销项目价格
int basket;
int special;
int dp[6][6][6][6][6]; int Decode (int c){
int i;
for (i=1; i<=5; ++i){
if (code[i] == c)
break;
}
return i;
} void Init(){
int i;
int j;
int c;
int k;
int index; scanf ("%d", &basket);
for (i=1; i<=basket; ++i){
scanf ("%d%d%d", &code[i], &num[i], &price[i]);
}
scanf ("%d", &special);
for (i=1; i<=special; ++i){
scanf ("%d", &special_cnt[i]);
for (j=1; j<=special_cnt[i]; ++j){
scanf ("%d%d", &c, &k);
index = Decode (c);
special_num[i][index] = k;
}
scanf ("%d", &special_price[i]);
}
} void Lowest_Price (){
int i1, i2, i3, i4, i5;
int i;
int tmp1, tmp2;
memset (dp, -1, sizeof(dp));
dp[0][0][0][0][0] = 0;
for (i1=0; i1<=num[1]; ++i1){
for (i2=0; i2<=num[2]; ++i2){
for (i3=0; i3<=num[3]; ++i3){
for (i4=0; i4<=num[4]; ++i4){
for (i5=0; i5<=num[5]; ++i5){
tmp1 = INT_MAX;
tmp2 = INT_MAX;
for (i=1; i<=special; ++i){
if (i1 >= special_num[i][1] &&
i2 >= special_num[i][2] &&
i3 >= special_num[i][3] &&
i4 >= special_num[i][4] &&
i5 >= special_num[i][5]){
tmp2 = dp[i1-special_num[i][1]]
[i2-special_num[i][2]]
[i3-special_num[i][3]]
[i4-special_num[i][4]]
[i5-special_num[i][5]] + special_price[i];
if (tmp1 > tmp2)
tmp1 = tmp2;
}
}
if (tmp1 != INT_MAX){
dp[i1][i2][i3][i4][i5] = tmp1;
}
else{
dp[i1][i2][i3][i4][i5] = i1 * price[1] + i2 * price[2]
+ i3 * price[3] + i4 * price[4] + i5 * price[5];
}
}
}
}
}
}
printf ("%d\n", dp[num[1]][num[2]][num[3]][num[4]][num[5]]);
} int main(void){
Init ();
Lowest_Price (); return 0;
}