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
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.
题意：给出N个商品，没给商品有利润pi和过期时间di，每天只能卖一个商品，过期商品不可以再卖，求如何安排卖每天的商品使得利润最大。
思路：
1.并查集：先排序商品的价值从大到小，然后从大到小开始扫描，如果过期那天的前一天没有的话则安排在那一天。、
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 1e4+5;
struct node
{
int p;
int d;
}a[N];
int parent[N];
bool cmp(node a,node b)
{
return a.p>b.p;
}
int ifind(int x)
{
if(parent[x]==-1)
return x;
return parent[x]=ifind(parent[x]);
}
int main()
{
int n;
while(scanf("%d",&n)==1){
memset(parent,-1,sizeof(parent));
for(int i=0;i<n;i++)
scanf("%d%d",&a[i].p,&a[i].d);
sort(a,a+n,cmp);
int sum=0;
for(int i=0;i<n;i++){
int fx=ifind(a[i].d);
if(fx>0){
parent[fx]=fx-1;
sum+=a[i].p;
}
}
printf("%d\n",sum);
}
return 0;
}
2.小根堆
贪心：对于每个时间t，我们应该保证在产品不过期的情况下，尽量卖出利润前t大的商品。
先排序从小到大过期时间，建立一个小根堆（节点权值为利润），若当前商品的过期时间t等于商品的个数，则说明在之前的安排下前t天已经安排了t个商品卖出，此时，若当前商品的利润大于堆顶则进行替换。若当前商品的过期时间t大于商品的个数直接插入堆内
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
#include <queue>
using namespace std;
typedef long long ll;
const int N = 1e4+5;
struct node
{
int val;
int time;
}a[N];
bool cmp(node a,node b)
{
return a.time<b.time;
}
int main()
{
int n;
while(scanf("%d",&n)==1){
int ans=0;
priority_queue<int,vector<int>,greater<int> >que;
for(int i=0;i<n;i++)
scanf("%d%d",&a[i].val,&a[i].time);
sort(a,a+n,cmp);
for(int i=0;i<n;i++){
if(que.size()<a[i].time){
que.push(a[i].val);
ans+=a[i].val;
}
else {
if(que.top()<a[i].val){
ans-=que.top();
ans+=a[i].val;
que.pop();
que.push(a[i].val);
}
}
}
cout<<ans<<endl;
}
return 0;
}
