1070. Mooncake (25)
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).
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers N (<=1000), the number of different kinds of mooncakes, and D (<=500 thousand tons), the maximum total demand of the market. Then the second line gives the positive inventory amounts (in thousand tons), and the third line gives the positive prices (in billion yuans) of N kinds of mooncakes. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print the maximum profit (in billion yuans) in one line, accurate up to 2 decimal places.
Sample Input:
3 200
180 150 100
7.5 7.2 4.5
Sample Output:
9.45
思路
简单的贪心算法，求出每种月饼的单价，按单价的从高到低顺序去满足需要的月饼数就可以得到最大利润。
注：当需求超过总月饼时，月饼的索引index应不大于N - 1，也就是说将所有月饼的利润加起来就行，之前没注意检测这一边界条件导致段错误。
代码
#include<iostream>
#include<iomanip>
#include<vector>
#include<algorithm>
using namespace std; class mooncake
{
public:
float amount;
float profit;
float price;
}; bool compare(mooncake a,mooncake b)
{
return a.price > b.price;
} int main()
{
int N;
float demand;
while(cin >> N >> demand)
{
vector<mooncake> cakes(N);
for(int i = ;i < N;i++)
{
cin >> cakes[i].amount;
}
for(int i = ;i < N;i++)
{
cin >> cakes[i].profit;
cakes[i].price = cakes[i].profit/cakes[i].amount;
}
sort(cakes.begin(),cakes.end(),compare);
int index = ;
float sum = ;
while(demand > && index <= N - ) //demand很大时index会越界，要注意判断下。
{
if(demand - cakes[index].amount > )
{
sum += cakes[index].profit;
}
else
{
sum += demand * cakes[index].price;
}
demand -= cakes[index++].amount;
}
cout << fixed << setprecision() << sum << endl;
}
}