POJ 1456——Supermarket——————【贪心+并查集优化】

时间：2015-10-26 12:12:36 阅读：250 评论：0 收藏：0 [点我收藏+]

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Supermarket

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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.
技术分享

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.

题目大意：给你n种食品，每种食品有pi和di分别表示该食品能卖pi元，保质期是di天。过了保质期就不能出售了，即不能获利了。每种食物需要花费一天出出售。问你采取最优出售顺序最多能获利多少钱。

解题思路：首先考虑贪心。我们将食物的价值从大到小排序。如果该食物在保质期当天可以出售（没有将其他食物安排在这天出售），那么就让它在保质期当天出售，为其他食物尽量匀出时间。如果不能在当天出售，我们考虑向前找第一个没有安排卖食物的时间（用标记数组，如果在当天安排卖食物，那么标记为true）。这种属于暴力查找。可以用并查集优化这种暴力查找，我们可以让并查集的父亲域表示该保质期食物可以在sets[x].pa这天出售。每次让sets[rootx].pa = rootx-1。表示目前将保质期为rootx的安排在rootx-1这天出售。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;
const int maxn = 1e5+200;
struct Product{
    int p,d;
}products[maxn];
struct Set{
    int pa;
}sets[maxn];
bool cmp(Product a,Product b){
    return a.p>b.p;
}
int Find(int x){
    if(x == sets[x].pa){
        return x;
    }
    int tmp = sets[x].pa;
    sets[x].pa = Find(tmp); //路径压缩
    return sets[x].pa;
}
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i = 1; i <= maxn-10;i++){
            sets[i].pa = i;
        }
        for(int i = 1; i <= n; ++i){
            scanf("%d%d",&products[i].p,&products[i].d);
        }
        sort(products+1,products+1+n,cmp);
        int sum = 0;
        for(int i = 1; i <= n;i++){
            int rootx = Find( products[i].d );
            if(rootx <= 0){
                continue;
            }
            sets[rootx].pa = rootx -1;  
            sum += products[i].p;
        }
        printf("%d\n",sum);
    }
    return 0;
}

POJ 1456——Supermarket——————【贪心+并查集优化】

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原文地址：http://www.cnblogs.com/chengsheng/p/4910610.html

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