Supermarket POJ - 1456 (贪心+并查集)

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.

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.

题意描述： 给出一个n然后再给出n组数，每组由p、d两个数组成，分别表示一件物品在1-d天中的价值为p，然后让你求出在这n组数中能产生的最大价值总和sum是多少。

解题思路：按照价值从大到小排序，价值相同按照时间从大到小排序，然后用一个f数组标记这一天是否被占用，首先在初始化f数组时，让f[i]=i;因为如果这一天被占用他只能去前一天，此时只需要让f[i]–;就可以了。

AC代码

#include<stdio.h>
#include<algorithm>
using namespace std;

struct node
{
	int  prod,day;
}p[10010];
 
 bool cmp(node a,node b)
 {
	if(a.prod!=b.prod)
 		return a.prod>b.prod;
 	return a.day>b.day; 
 }
 int f[10010];
 
 void inti(int n)
 {
 	for(int i=0;i<=n;i++)
 		f[i]=i;
 	return ;
 }
 
 int merge(int x)
 {
 	if(f[x]==x)
 		return x;
	 f[x]=merge(f[x]);
	 return f[x];
 }
 
 int main(void)
 {
 	int i,j,n,tx,maxx;
	long long sum;
 	while(~scanf("%d",&n)){
 		sum=0;maxx=0;
 		for( i=0;i<n;i++){
 			scanf("%d%d",&p[i].prod,&p[i].day);
 			maxx=max(maxx,p[i].day);
		 }
 		sort(p,p+n,cmp);
 		inti(maxx);
 		for(i=0;i<n;i++){
 			tx=merge(p[i].day);
 			if(tx>0){//因为天数必须大于0
 				f[tx]--;//表示这一天被占用，只能去前一天了。 
 				sum+=p[i].prod;
			 }
		 }
 		printf("%lld\n",sum);
	 }
	 return 0;
 }