sicily 2000. Toy Shopping

Description
Bessie wants some toys. She's been saving her allowance for years, and has an incredibly huge stash. However, she is quite frugal and wants to get the best value for her cash. In fact, she has decided only to buy exactly three different toys of the N (3 <= N <= 25,000) offered at the Bovine Plaything Palace.
Toy i brings Bessie J_i (0 <= J_i <= 1,000,000) microbundles of joy and and has price P_i (0 < P_i <= 100,000,000). Bessie has enough money to buy any three toys that she chooses.
Bessie wants to maximize the sum of her happy-frugal metric (which is calculated as J_i/P_i -- joy divided by price) for the three toys she chooses. Help Bessie decide which toys she should buy. The answer is guaranteed to be unique.
Assume that the Bovine Plaything Palace offers 6 different toys for Bessie:

        i    Joy       Price       Happy-Frugal Metric

        -    ---       -----       -------------------

        1      0        521               0.00000

        2    442        210               2.10476...

        3    119        100               1.19000

        4    120        108               1.11111...

        5    619        744               0.83198...

        6     48         10               4.80000


Bessie would choose toy 6 (HFM = 4.80), toy 2 (HFM = 2.10), and toy 3 (HFM = 1.19).

Input
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains two space-separated integers: J_i and P_i

Output
* Line 1: The total price that Bessie will have to pay
* Lines 2..4: In descending order sorted by the happy-frugal metric, the 1-based index of the toys that Bessie should buy, one per line

再做题难度大点的体会下C++的新特性，包括结构体使用时可以省略 struct，在for的head里声明循环变量使程序结构更清晰，还有sort函数的使用
思路很简单，输入，排序，再输出前三，用时0.06s，不是很好

#include<iostream>
#include <algorithm>
using namespace std;

struct toy
{
    double hfm;
    int index, joy, price;
} a[26000];

bool cmp( toy a, toy b )
{
    return a.hfm > b.hfm;
}
int main()
{
    int n;
    cin >> n;
    
    for( int i = 0; i < n; i++ )
    {
        cin >> a[i].joy >> a[i].price;
        a[i].index = i+1;
        a[i].hfm = a[i].joy * 1.0 / a[i].price;
    }
    
    sort( a, a + n, cmp );
    
    int sum = 0;
    for ( int i = 0; i < 3; i++ )
        sum += a[i].price;
    cout << sum << endl;
    
    for ( int i = 0; i < 3; i++ )
        cout << a[i].index << endl;
}