Supermarket -- 贪心 + 并查集

题目：

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

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题意分析：

输入n代表有n件物品，每件物品具有2个属性，第1个属性是他的利润，在规定时间内卖出可以获得该利润，第2个属性是他的规定日期，意思是可以在给出的这个日期之前卖出该物品。

思路：

用结构体数组存所有的物品，然后按照利润排序，同时用pre数组存卖出的时间。排完序之后，所有的物品是按照利润从大到小排列的，每次循环取出，最大利润的物品在规定期限内售卖，如果这一天卖出了这件物品的话，那么同日期的物品都要向前提前一天即：pre[j] = j - 1;这样把以最大的收益在规定时间内卖出所有的物品。

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代码如下：

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;int pre[10005];int find_boss(int x){    if(pre[x] != x)        pre[x] = find_boss(pre[x]);    return pre[x];}struct product{    int value;    int date;}ar[10005];void init(int n){    for(int i = 0 ; i<= n; i++)        pre[i] = i;}bool cmp(product a, product b){    return a.value > b.value;}int main(){    int n;    while(cin >> n)    {        init(10005);        for(int i = 0; i < n; i++)        {            cin >> ar[i].value >> ar[i].date;        }        int ans = 0;        sort(ar, ar + n, cmp);        for(int i = 0 ; i < n; i++)        {            int j = find_boss(ar[i].date);            if(j > 0)            {                ans += ar[i].value;                pre[j] = j - 1;            }        }        cout << ans << endl;    }    return 0;}

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