USACO: Shopping Offers

Shopping Offers
IOI'95

In a certain shop, each kind of product has an integer price. For example, the price of a flower is 2 zorkmids (z) and the price of a vase is 5z. In order to attract more customers, the shop introduces some special offers.

A special offer consists of one or more product items together for a reduced price, also an integer. Examples:

• three flowers for 5z instead of 6z, or
• two vases together with one flower for 10z instead of 12z.

Write a program that calculates the price a customer has to pay for a purchase, making optimal use of the special offers to make the price 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 14z: two vases and one flower for the reduced price of 10z and two flowers for the regular price of 4z.

PROGRAM NAME: shopping

INPUT FORMAT

The input file has a set of offers followed by a purchase.

Line 1:	s, the number of special offers, (0 <= s <= 99).
Line 2..s+1:	Each line describes an offer using several integers. The first integer is n (1 <= n <= 5), the number of products that are offered. The subsequent n pairs of integers c and k indicate that k items (1 <= k <= 5) with product code c (1 <= c <= 999) are part of 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.
Line s+2:	The first line contains the number b (0 <= b <= 5) of different kinds of products to be purchased.
Line s+3..s+b+2:	Each of the subsequent 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 to be purchased (1 <= k <= 5). The value p is the regular price per item (1 <= p <= 999). At most 5*5=25 items can be in the basket.

SAMPLE INPUT (file shopping.in)

2
1 7 3 5
2 7 1 8 2 10
2
7 3 2
8 2 5

OUTPUT FORMAT

A single line with one integer: the lowest possible price to be paid for the purchases.

SAMPLE OUTPUT (file shopping.out)

14

 

思路：

DP，挺简单的，用一个五维数组保存各种组合，当然，调试的过程要花点时间去想。我在中间加了一点优化，就是先检测一下各组优惠组合对于我们要买的这些东西是否真的能用得上，不用的话就直接忽略掉这个组合。然后O(5^5*99)的检测。

看起来我的这个优化起到一点作用——中间几组数据挺大的反而0s过。

 

还看到官方的分析有个挺NB的——转换成图论，每种组合看做一个节点，假设要买5种物品各abcde个，就是求V₀₀₀₀₀到V_abcde的最短路，这么好的想法真是挺让我震惊的！

/* ID: geterns1 PROG: shopping LANG: C++ */ #include <iostream> #include <fstream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; int items[1000]; int s, offers[100][5], thNum[100][5], rPrice[100], Count[100]; int b, itNum[5], itPrice[5]; int rs, spNum[100][5], spPrice[100]; int best[6][6][6][6][6]; int main(){ freopen("shopping.in", "r", stdin); //重定向输入 freopen("shopping.out", "w", stdout);//重定向输出 // scanf("%d", &s); for (int i = 0; i < s; i ++){ scanf("%d", Count + i); for (int j = 0; j < Count[i]; j ++) scanf("%d %d", &offers[i][j], &thNum[i][j]); scanf("%d", rPrice + i); } // scanf("%d", &b); int tmp; for (int i = 0; i < 1000; i ++) items[i] = -1; for (int i = 0; i < b; i ++){ scanf("%d %d %d", &tmp, itNum + i, itPrice + i); items[tmp] = i; } // int map[5], rs = 0; bool nouse; for (int i = 0; i < s; i ++){ nouse = false; memset(map, 0, sizeof(map)); for (int j = 0; j < Count[i]; j ++){ if (items[offers[i][j]] < 0){ nouse = true; break; } map[items[offers[i][j]]] = thNum[i][j]; } if (nouse) continue; for (int j = 0; j < 5; j ++){ if (map[j] > itNum[j]){ nouse = true; break; } spNum[rs][j] = map[j]; } if (nouse) continue; spPrice[rs ++] = rPrice[i]; } // int minCost; int pduct[5][6]; for (int i = 0; i < 5; i ++) for (int j = 0; j < 6; j ++) pduct[i][j] = itPrice[i] * j; for (int a = 0; a <= itNum[0]; a ++){ for (int b = 0; b <= itNum[1]; b ++){ for (int c = 0; c <= itNum[2]; c ++){ for (int d = 0; d <= itNum[3]; d ++){ for (int e = 0; e <= itNum[4]; e ++){ minCost = pduct[0][a] + pduct[1][b] + pduct[2][c] + pduct[3][d] + pduct[4][e]; for (int i = 0; i < rs; i ++){ if (a >= spNum[i][0] && b >= spNum[i][1] && c >= spNum[i][2] && d >= spNum[i][3] && e >= spNum[i][4]){ if (best[a - spNum[i][0]][b - spNum[i][1]][c - spNum[i][2]][d - spNum[i][3]][e - spNum[i][4]] + spPrice[i] < minCost) minCost = best[a - spNum[i][0]][b - spNum[i][1]][c - spNum[i][2]][d - spNum[i][3]][e - spNum[i][4]] + spPrice[i]; } } best[a][b][c][d][e] = minCost; } } } } } printf("%d/n", best[itNum[0]][itNum[1]][itNum[2]][itNum[3]][itNum[4]]); fclose(stdin); //关闭文件 fclose(stdout); //关闭文件 return 0; }