据说这题是考查的并查集（表示完全不会。。），但被我用贪心（这次应该是贪心。。）给AC了，第一次提交TLE了，显然是排序用的冒泡没用快排，不得不搜集资料找快排函数用法，于是顺便会用快排函数了
Description
A supermarket has a set Prod of products on sale. It earns a profit px for each product x∈Prod sold by a deadline dx that is measured as an integral number of time units starting from the moment the sale begins. Each product takes precisely one unit of time for being sold. A selling schedule is an ordered subset of products Sell ≤ Prod such that the selling of each product x∈Sell, according to the ordering of Sell, completes before the deadline dx or just when dx expires. The profit of the selling schedule is Profit(Sell)=Σ
x∈Sellpx. An optimal selling schedule is a schedule with a maximum profit.
For example, consider the products Prod={a,b,c,d} with (pa,da)=(50,2), (pb,db)=(10,1), (pc,dc)=(20,2), and (pd,dd)=(30,1). The possible selling schedules are listed in table 1. For instance, the schedule Sell={d,a} shows that the selling of product d starts at time 0 and ends at time 1, while the selling of product a starts at time 1 and ends at time 2. Each of these products is sold by its deadline. Sell is the optimal schedule and its profit is 80.
Write a program that reads sets of products from an input text file and computes the profit of an optimal selling schedule for each set of products.
Input
A set of products starts with an integer 0
Output
For each set of products, the program prints on the standard output the profit of an optimal selling schedule for the set. Each result is printed from the beginning of a separate line.
Sample Input
4 50 2 10 1 20 2 30 17 20 1 2 1 10 3 100 2 8 2
5 20 50 10
Sample Output
80185
Hint
The sample input contains two product sets. The first set encodes the products from table 1. The second set is for 7 products. The profit of an optimal schedule for these products is 185.
#include
#include
#include
#include
#include
#include
using namespace std;
struct node
{
int p, d;
} fei[10001];
int a[10001];
bool operator b.p;
}
int main()
{
int n, i, j, s, x;
while(scanf("%d",&n)!=EOF)
{
s = 0;
memset(a,0,sizeof(a));
for(i=0; i=1; j--)
{
if(a[j]==0)
{
s=s+fei[i].p;
a[j]++;
break;
}
}
}
}
printf("%dn",s);
}
return 0;
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