POJ 1456 Supermarket 并查集（想不到的思路）



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 2   5 20  50 10

Sample Output

80185

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：就是给定n 种商品，给定每种商品的最后销售期限，求在最后期限能够得到的最大价值；

思路：刚开始以为只要求的各个相同期限的最大值相加就可以了，但后来发现看到给的表格里有a，c；就说明在第二个期限里可以售出期限大的且值大于当前期限的值得商品

看来别人的博客，用并查集可以计算：首先对给定的数据进行值排序，然后用并查集的思想，让最小的为父节点，看代码说吧，一目了然..

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;
struct point{
    int x;
    int endline;
}p[10005];
int cmpx(point a,point b){
    return a.x > b.x;
}//排序
int fu[10005];

int found(int x){
    return x = x == fu[x] ? x : found(fu[x]);
}//找其父节点

int main(){
    int n;
    while(~scanf("%d",&n)){
        for(int i = 0;i < 10005;i++){
            fu[i] = i;
        }
        for(int i = 0;i < n; i++){
            scanf("%d%d",&p[i].x,&p[i].endline);
        }
        sort(p,p+n,cmpx);//排序
        int sum = 0;
        for(int i = 0;i < n; i++){
            int t = found(p[i].endline);//找到期限的父亲，大于0就进行累加；
            if(t > 0){
                fu[t] = t - 1;//把父亲节点的值减小；
                sum += p[i].x;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}

注意：这题就是思路想的不太好，譬如4 ，50 2 ，10 1，20 2，30,1；排序后变为 50（2），30（1），20（2），10（1）；执行过程，f[2] = 2变为1；f[1] = 1变为0，第三次到20这个值得时候，他的父亲节点找到为0；所以不累加；好难得思想。继续努力。。。。。。