POJ1456：Supermarket（并查集+贪心）

Posted 2021-01-21 heyuhhh

Supermarket

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17634		Accepted: 7920

题目链接：http://poj.org/problem?id=1456

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

题意：

给出n个商品以及它们的利润和截至日期，每天只能卖出一个商品，截止日期过后就不能出售了，问最大利润是多少。

 

题解：

第一反应肯定是采用贪心的策略，选择一个最大利润的出售，但是因为这里有个日期，这里的贪心不一定可以得到最后的最大利润。

所以我们也应该考虑日期，发现当出售商品在其截至日期时，是最优的，因为这将不会影响之前商品的出售。

这里的贪心我们可以采用优先队列对其进行优化，并且日期要倒过来枚举（结合贪心策略想想）。

这里我用的是并查集对其进行优化，商品出售的时间尽可能得接近其截至日期，如果在那天商品出售了，又有另外一个商品在那天出售，那么我们就挪到前一天进行出售。

我们就用并查集维护离截止日期最近的空闲时间，查找速度挺快的。

 

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

const int N = 10005; 
int f[N];
int n;

struct node{
    int p,d;
    bool operator < (const node &A)const{
        return A.p<p;
    }
}a[N];
int find(int x){return f[x]==x ? f[x] : f[x]=find(f[x]);}
int main(){
    while(~scanf("%d",&n)){
        for(int i=1;i<=10000;i++) f[i]=i;
        for(int i=1;i<=n;i++) scanf("%d%d",&a[i].p,&a[i].d);
        sort(a+1,a+n+1);
        int ans = 0;
        for(int i=1;i<=n;i++){
            int fd = find(a[i].d);
            if(fd>0){//存在空闲时间
                ans+=a[i].p;
                f[fd]=fd-1;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}