题干:
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 17 20 1 2 1 10 3 100 2 8 25 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:
超市里有N个商品. 第i个商品必须在保质期(第di天)之前卖掉, 若卖掉可让超市获得pi的利润.
每天只能卖一个商品.
现在你要让超市获得最大的利润。
解题报告:
优先队列就行了,考虑成每一个cur天的决策,如果这一天还不到截止日期,那就直接push,并且把cur++。如果到了截止日期,那就只能看看需不需要更新pq中的元素,最后元素的个数一定就是答案。这题还有并查集做法:【51Nod - 1163】最高的奖励,按权值排序,遍历来找到可以最早安放的日期。
AC代码:
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#include<bitset>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> pii;
const int MAX = 2e5 + 5;
int n;
pii p[MAX];
int main()
{while(~scanf("%d",&n)) {for(int i = 1; i<=n; i++) cin>>p[i].second>>p[i].first;sort(p+1,p+n+1);priority_queue<int,vector<int>,greater<int> > pq;int cur = 1;for(int i = 1; i<=n; i++) {if(cur <= p[i].first) {pq.push(p[i].second);cur++;}else {if(p[i].second > pq.top()) pq.pop(),pq.push(p[i].second);}} int ans = 0;while(!pq.empty()) ans += pq.top(),pq.pop();printf("%d\n",ans);}return 0 ;
}