2000. Toy Shopping 继续是struct

2000. Toy Shopping

Constraints

Time Limit: 1 secs, Memory Limit: 64 MB

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

Sample Input

60 521442 210119 100120 108619 74448 10

Sample Output

320623

Problem Source

2010中山大学新手赛-网络预选赛

第一次先用了冒泡排序 结果在第10个case的时候TLE了 

代码如下：

#include <iostream>using namespace std;struct Toy {int num;int joy;int price;double HFM;}toy[25001];int main () {int n;cin>>n;for (int i = 1; i <= n; i++) { //输入toy[i].num = i;cin>>toy[i].joy;cin>>toy[i].price;toy[i].HFM = (double)toy[i].joy / (double)toy[i].price;}Toy temp;for (int i = 1; i <= n; i++) {  //冒泡for (int j = 1; j <= n - i; j++) {if (toy[j].HFM < toy[j+1].HFM) {temp = toy[j];toy[j] = toy[j+1];toy[j+1] = temp;}}}int count = 0;for (int i = 1; i <= 3; i++) {  //计算价格count += toy[i].price;}cout<<count<<endl;for (int i = 1; i <= 3; i++) {cout<<toy[i].num<<endl;}//system("pause");return 0;}

后来去掉了冒泡，直接用一个数组记录HFM最大的3个toy的下标，然后直接输出就AC了。

代码如下：

#include <iostream>using namespace std;struct Toy {int num;int joy;int price;double HFM;}toy[25001];int main () {int n;cin>>n;for (int i = 1; i <= n; i++) {toy[i].num = i;cin>>toy[i].joy;cin>>toy[i].price;toy[i].HFM = (double)toy[i].joy / (double)toy[i].price;}int count = 0;int a[4] = {0};int k = 0;while (k < 3) {double max = toy[0].HFM;for (int i = 1; i <= n; i++) {if (max < toy[i].HFM) {max = toy[i].HFM;a[k] = i;}}toy[a[k]].HFM = 0;k++;}for (int i = 1; i <= 3; i++) {count += toy[a[i - 1]].price;}cout<<count<<endl;for (int i = 1; i <= 3; i++) {cout<<toy[a[i - 1]].num<<endl;}//system("pause");return 0;}