Supermarket

Time Limit: 2000MS	Memory Limit: 65536K
Total Submissions: 7677	Accepted: 3252

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

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. 題意：有n個商品，每個商品都有相應的價值pi和賣出去的最後期限di，每天只能賣一件商品。求最多能獲得的價值。 思路：貪心法。 法一：並查集優化：列舉每個商品，找出最接近該商品最後期限的未被佔用的一天來賣該商品。而要找出這一天可以用到並查集來快速查詢。 AC程式碼：

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdlib>

using namespace std;
struct node
{
    int pi,di;
}pro[10005];
int fa[10005];
bool cmp(node a,node b)
{
    return a.pi>b.pi;
}
void init()
{
    for(int i=1;i<=10005;i++)
    fa[i]=i;
}
int Find_set(int x)
{
    if(x==fa[x]) return x;
    fa[x]=Find_set(fa[x]);
    return fa[x];
}
int main()
{
   int n;
   while(scanf("%d",&n)!=EOF)
   {
       for(int i=0;i<n;i++)
       scanf("%d%d",&pro[i].pi,&pro[i].di);
       sort(pro,pro+n,cmp);
       init();
       int sum=0;
       for(int i=0;i<n;i++)
       {
           int rt=Find_set(pro[i].di);
           if(rt>0)
           {
               sum+=pro[i].pi;
               fa[rt]=rt-1;
           }
       }
       printf("%d\n",sum);
   }
   return 0;
}

法二：優先佇列優化 AC程式碼：

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdlib>

using namespace std;
struct node
{
    int pi,di;
}pro[10005];
int fa[10005];
bool cmp(const node& p1,const node& p2){
	return p1.di<p2.di;
}
bool operator < (const node& p1,const node& p2){
	return p2.pi<p1.pi;
}
int main()
{
   int n;
   priority_queue<node>q;
   while(scanf("%d",&n)!=EOF)
   {
       for(int i=0;i<n;i++)
       scanf("%d%d",&pro[i].pi,&pro[i].di);
       sort(pro,pro+n,cmp);
       int sum=0;
       for(int i=0;i<n;i++)
       {
           if(pro[i].di>q.size())
           {
               q.push(pro[i]);
           }
           else if(pro[i].pi>q.top().pi)
           {
               q.pop();
               q.push(pro[i]);
           }
       }
       while(!q.empty())
       {
           sum+=q.top().pi;
           q.pop();
       }
       printf("%d\n",sum);
   }
   return 0;
}