Atcoder C - Half and Half

比萨订购问题

最新推荐文章于 2019-01-25 13:49:00 发布

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本文介绍了一个关于订购不同种类比萨的算法问题。通过使用贪心算法来确定如何以最低成本获得所需的A比萨和B比萨数量。文章提供了一段C++代码实现，并解释了算法的逻辑。

C - Half and Half

Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively.

Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

Constraints

1≤A,B,C≤5000
1≤X,Y≤105
All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C X Y

Output

Print the minimum amount of money required to prepare X A-pizzas and Y B-pizzas.

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#define LL long long
using namespace std;
int pra, prb, prc;
int x, y;
LL ans = 0;
int main()
{
	while (cin >> pra >> prb >> prc >> x >> y) {
		ans = 0;
		if (prc * 2 < pra + prb) {
			ans += min(x, y)*prc * 2;
			if (x > y)
				ans += (x - y)*pra;
			else ans += (y - x)*prb;
		} else {
			ans = x*pra + y*prb;
		}
		if (prc * 2 * max(x, y) < ans)
			ans = prc * 2 * max(x, y);
		cout << ans << endl;
	}
	return 0;
}

贪心算法，如果存在C的价格乘2低于A+B，对于A和B都要买的情况，全部选择C来进行购买。之后再分类讨论就OK了。