Supermarket

title: Supermarket
tags: [贪心,并查集]

题目链接

描述

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)=Σx∈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.

输入

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.

输出

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.

样例输入

4  50 2  10 1   20 2   30 1

7  20 1   2 1   10 3  100 2   8 2

   5 20  50 10

样例输出

80
185

/**
贪心思想加并查集，以前做的纯贪心，这里提交会超时，
按照利益从大到小排完序后，遍历每个物品，在他的最后期限（d）的这一天卖掉，
按照题意，在d天就不能再卖其他东西了，若是后面的物品的最后期限也是在d天，
那么这件物品只能在d-1天卖掉。即把这件物品的最后期限改为d-1


*/

#include<bits/stdc++.h>
using namespace std;
struct node
{
    int pro;
    int date;
};
bool cmp(node a,node b)
{
    return a.pro>b.pro;
}
int vest[10010];
void init()
{
    for(int i=0; i<=10010; i++)
        vest[i]=i;
}
int findx(int t)
{
    if(vest[t]==t)return t;
    else return vest[t]=findx(vest[t]);
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        node goods[10010];
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&goods[i].pro,&goods[i].date);
        }
        sort(goods,goods+n,cmp);
        int sum=0;
        init();
        for(int i=0; i<n; i++)
        {
            int a=findx(goods[i].date);///此物品的最后期限
            if(a!=0)///此物品的最后期限不是0
            {
                sum+=goods[i].pro;
                vest[a]=findx(a-1);///以后若还有商品的最后期限为a的话，那么就将他的最后期限向前移动一天
            }
        }
        printf("%d
",sum);
    }
    return 0;
}