zoukankan      html  css  js  c++  java
  • poj 1170 Shopping Offers

    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

    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.

    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

    Source

    此题用到6进制表示物品。吧套餐也看作是物品,这里用6进制将他们区分开。然后就老实背包。
    #include <stdio.h>
    #include <string.h>
    #pragma warning(disable:4996)
    int b,c,k,p,s,n, a[200], v[200],w[200];
    int dp[200000+20],id[1000],sum;
    int main()
    {
        while (~scanf("%d", &b))
        {
            int six = 1;
            sum = 0;
            memset(id, -1, sizeof id);
            for (int i = 0; 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 *= 6;
            }
            for (int i = 1; i <= sum; i++)dp[i] = 1 << 28;
            scanf("%d", &s);
            while (s--)
            {
                a[b] = 1<<28;
                w[b] = 0;
                scanf("%d", &n);
                int s = 0;
                for (int i = 0; i < n; i++)
                { 
                    scanf("%d%d", &c, &k);
                    if (id[c] == -1)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] == 1<<28)a[b] = 0;
                scanf("%d", &p);
                v[b] = s<p?s:p;
                b++;
            }
            for (int i = 0; i < b; i++)
                for (int k = 0; 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
    ", dp[sum]);
            //for (int i = 0; i <= sum; i++)    printf("%d
    ", dp[i]);
        }
    }
  • 相关阅读:
    hadoop2.2.0 centos6.4 编译安装详解
    Hadoop 2.2.0的高可用性集群中遇到的一些问题(64位)
    Visual Studio 常用快捷键
    Android(1)—Mono For Android 环境搭建及破解
    IbatisNet SqlMap.config配置节导致的程序无法通过
    CAD数据分块,偏移校准,加载到百度地图、高德地图、谷歌等地图上
    数据库SQL优化大总结
    Scratch 3下载,最新版Scratch下载,macOS、Windows版
    高性能网站设计之缓存更新的套路
    【验证无效】MySQL的count(*)的优化,获取千万级数据表的总行数
  • 原文地址:https://www.cnblogs.com/newadi/p/4750885.html
Copyright © 2011-2022 走看看