Supermarket POJ

Supermarket

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.

 

 

题目：每个物品有一个价值px，保质期dx，在这保质期之前都可以出售。一天只能出售一个物品，问出售这些物品价值总和最大是多少。

思路：我们可以把物品按照价值排序，我们知道，这题可以贪心，从最大的价值开始判断可不可以取。我们需要快速查询该物品保质期之前最近的哪一天，在那天购买该物品。我们可以用并查集来进行查询操作，我们进行fa[i] = i。

rt = Find(i)

① fa[rt] != 0 则说明前面有一天没有买东西，则我们把fa[rt] = rt - 1，说明我们把那天使用。这样我们也可以串联起零散的并查集树。

② fa[rt] == 0,说明前面所有天都被使用，则无法购买该物品。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>

using namespace std;

#define ll long long
#define pb push_back
#define fi first
#define se second

const int N = 10010;
struct node
{
    int id, v;

    bool friend operator<(const node& a, const node& b){
        return a.v > b.v;
    }
};
int fa[N];

int Find(int x){
    return fa[x] == x ? x : fa[x] = Find(fa[x]);
}

void solve()
{   
    int n;
    while(~scanf("%d", &n)){ 
        vector<node > vn;
        
        for(int i = 1; i <= n; ++i){
            int id, v;
            scanf("%d%d", &v, &id);
            vn.pb({id, v});
        }

        sort(vn.begin(), vn.end());
        // for(auto vv : vn) cout << vv.v << " ";
        // cout << endl;
        for(int i = 0; i < N; ++i) fa[i] = i;

        int profits = 0;
        for(int i = 0; i < n; ++i){
            int rt = Find(vn[i].id);    
            if(rt != 0){
                profits += vn[i].v;
                fa[rt] = rt - 1;
            }
        }

        //printf("profits = %d
", profits);
        printf("%d
", profits);
    }
}

int main()
{

    solve();

    return 0;
}