POJ 1170 Shopping Offers -- 动态规划(虐心的六重循环啊！！！)

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

Output

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;
}

POJ 1170 Shopping Offers -- 动态规划(虐心的六重循环啊！！！)

原文：http://blog.csdn.net/jdplus/article/details/19827909