PAT 解题报告 1070. Mooncake (25)

1070. Mooncake题目描述:

Mooncake is a Chinese bakery product traditionally eaten during the Mid-Autumn Festival. Many types of fillings and crusts can be found in traditional mooncakes according to the region’s culture. Now given the inventory amounts and the prices of all kinds of the mooncakes, together with the maximum total demand of the market, you are supposed to tell the maximum profit that can be made.

Note: partial inventory storage can be taken. The sample shows the following situation: given three kinds of mooncakes with inventory amounts being 180, 150, and 100 thousand tons, and the prices being 7.5, 7.2, and 4.5 billion yuans. If the market demand can be at most 200 thousand tons, the best we can do is to sell 150 thousand tons of the second kind of mooncake, and 50 thousand tons of the third kind. Hence the total profit is 7.2 + 4.5/2 = 9.45 (billion yuans).

就是说有一系列的月饼, 包括月饼的数量, 和数量对应的销售金额, 同时给出市场需求的月饼数量, 求最大的利润.

1070. Mooncake算法分析:

这个是贪心算法. 求得每种月饼的单位数量上的售价, 按照这个单位售价降序排列, 依次尽量取得单位售价比较高的去填充那个市场需求, 知道达到那个最大的市场需求. 所得到的利润就是最大的. 换个角度就说市场的最大需求量固定, 那么我们当然希望这些个需求量里面单位数量的销售价格越高越好, 所以我们降序排列就可以得到最大利润. AC代码如下:

#include <cstdio>
#include <algorithm>
#include <vector>

struct Node {
    double amount;
    double price;

bool cmp(const Node &n1, const Node &n2) {
    return n1.price * n2.amount > n2.price * n1.amount;

double gao(int n, double d, std::vector<Node> &cakes) {
    std::sort(cakes.begin(), cakes.end(), cmp);
    double profit = 0.0;
    for(int i = 0; d > 0 && i < n; ++i) {
        int sell = std::min(d, cakes[i].amount);
        profit += cakes[i].price * (sell * 1.0 / cakes[i].amount);
        d -= sell;
    return profit;

int main() {
    int N, D;
    scanf("%d %d", &N, &D);

    std::vector<Node> cakes(N);
    for(int i = 0; i < N; ++i)
        scanf("%lf", &cakes[i].amount);
    for(int i = 0; i < N; ++i)
        scanf("%lf", &cakes[i].price);

    printf( "%.2lf\n", gao(N, D, cakes) );
    return 0;

1070. Mooncake注意点:

有个注意点就是每种月饼的数量需要用浮点数保存, 因为测试用例里面这个数量不一定是整型的.

Written on November 30, 2013