POJ 1456 Supermarket【贪心+并查集】

Supermarket

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11450		Accepted: 5053

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 <= n <= 10000, which is the number of products in the set, and continues with n pairs pi di of integers, 1 <= pi <= 10000 and 1 <= di <= 10000, that designate the profit and the selling deadline of the i-th product. White spaces can occur freely in input. Input data terminate with an end of file and are guaranteed correct.

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 1

7  20 1   2 1   10 3  100 2   8 2
   5 20  50 10

Sample Output

80
185

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.

Source

Southeastern Europe 2003

题目大意：有n个商品，每个商品都有一个值，第一个是其价值，第二个是其过期日期，每一天只能卖一件商品，问最大利润。

思路：

按照商品价值从高到低排序，然后从高到低遍历，如果当前商品的截止日期没有分配任务，那么就在这个日期上分配当前商品任务，如果截止日期上已经分配了商品任务了，就向前遍历日期，看哪一天空闲，然后在这个空闲的日期上分配当前商品任务，如果从当前商品任务的截止日期到第一天都有商品任务分配了，那么就说明这个商品任务没有必要完成了。

网上看到有很多大牛用并查集来优化问题，尝试了一发，的确是快了很多，具体操作为这样：

初始的时候，将每个f【】都设定自身为根节点，在遍历当前任务的时候，令ff=find（a【i】.day），如果ff>0那么设定上这个日期分配当前任务，并且使得f【ff】=ff-1，这样就达成了上述最开始思路的操作了。

AC代码：

#include
#include
#include
#include
using namespace std;
struct node
{
    int val,day;
}a[100000];
int f[100000];
int cmp(node a,node b)
{
    return a.val>b.val;
}
int find(int a)
{
    int r=a;
    while(f[r]!=r)
    r=f[r];
    int i=a;
    int j;
    while(i!=r)
    {
        j=f[i];
        f[i]=r;
        i=j;
    }
    return r;
}
void merge(int a,int b)
{
    int A,B;
    A=find(a);
    B=find(b);
    if(A!=B)
    f[B]=A;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int maxday=0;
        for(int i=0;i0)
            {
                f[ff]=ff-1;
                output+=a[i].val;
            }
        }
        printf("%d\n",output);
    }
}