PAT 1037. Magic Coupon (25)

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本文介绍了一种在火星魔法店利用正负值优惠券和商品价值获取最大收益的方法。通过将数据分为同号存储并利用数学原理进行最优匹配，实现了算法设计。

The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product back! What is more, the shop also offers some bonus product for free. However, if you apply a coupon with a positive N to this bonus product, you will have to pay the shop N times the value of the bonus product... but hey, magically, they have some coupons with negative N's!

For example, given a set of coupons {1 2 4 -1}, and a set of product values {7 6 -2 -3} (in Mars dollars M$) where a negative value corresponds to a bonus product. You can apply coupon 3 (with N being 4) to product 1 (with value M$7) to get M$28 back; coupon 2 to product 2 to get M$12 back; and coupon 4 to product 4 to get M$3 back. On the other hand, if you apply coupon 3 to product 4, you will have to pay M$12 to the shop.

Each coupon and each product may be selected at most once. Your task is to get as much money back as possible.

Input Specification:

Each input file contains one test case. For each case, the first line contains the number of coupons NC, followed by a line with NC coupon integers. Then the next line contains the number of products NP, followed by a line with NP product values. Here 1<= NC, NP <= 10⁵, and it is guaranteed that all the numbers will not exceed 2³⁰.

Output Specification:

For each test case, simply print in a line the maximum amount of money you can get back.

Sample Input:

Sample Output:

/* 
 /*  1.此题将数据分为同号的存储即可。 
     2.还有如果a>b,c>d 那么 a*c+b*d > a*d+b*c的（很好证。。）
	 基于以上2点 很容易设计算法了。。
*/
#include "iostream"
#include "vector"
#include "cstring"
#include "string"
#include "stack"
#include "algorithm"
using namespace std;
bool cmp1(int a,int b){
	return a < b;
}
bool cmp2(int a, int b) {
	return a > b;
}
int main() {
	int n;
	vector<int>v1, v2, v3, v4;
	cin >> n;
	for (int i = 0; i < n; i++) {
		int t;
		cin >> t;
		if (t > 0) {
			v1.push_back(t);
		}
		else {
			v2.push_back(t);
		}
	}
	int m;
	cin >> m;
	for (int i = 0; i < m; i++) {
		int t;
		cin >> t;
		if (t > 0) {
			v3.push_back(t);
		}
		else {
			v4.push_back(t);
		}
	}
	sort(v1.begin(), v1.end(),cmp2);
	sort(v2.begin(), v2.end(), cmp1);
	sort(v3.begin(), v3.end(), cmp2);
	sort(v4.begin(), v4.end(), cmp1);
	long long sum = 0;
	int i = 0;
	int j = 0;
	while (i < v1.size() && j < v3.size()) {
		sum += v1[i] * v3[j];
		i++; j++;
	}
	i = 0; j = 0;
	while (i < v2.size() && j < v4.size()) {
		sum += v2[i] * v4[j];
		i++; j++;
	}
	cout << sum << endl;
	return 0;
}