POJ 1456 supermarket（贪心加并查集）

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.

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.
题目大意：
超市里有n个产品要卖，每个产品都有一个截至时间dx（从开始卖时算起），只有在这个截至时间之前才能卖出并且获得率润dy。
有多个产品，所有可以有不同的卖出顺序，每卖一个产品要占用1个单位的时间，问最多能卖出多少利润。

1 仅贪心解决，按照价值从大到小进行排列，每次取最大的价值，从他结束的时间开始向前暴力，如果有时间可以使用，即没被vis数组标记，就可加上此位置的价值
2 贪心用并查集进行优化，还是按照价值问题进行排列，每次回溯，找前面有一个点pre【i】==i，那么此点可以使用，就可以停止进行，此外此点自己- -，表示该时间被使用，后面如果回溯到该点的时候不可以停止下来，仍然需要向前回溯，如果一直回溯到0，即没有位置可以进行，那么这次的价值也不会算到总和里面

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<vector>using namespace std;#define MAXN 10010int pre[MAXN];bool vis[MAXN];int find(int x){    int r = x;      //用r来传递信息    while(pre[r] != r)  //如果r的上级bushir自身        r = pre[r];     //r就接着找上级，直到找到老大    return r;           //告诉你x的老大是谁}struct money{    int x, y;    friend bool operator < (const money &a, const money &b){        return a.x > b.x;    }}arr[MAXN];int main(){    int n, m, i, j, a, b;    while(cin >> n)    {        int maxtime = 0, sum = 0;        for(i=0; i<n; i++){            cin >> arr[i].x >> arr[i].y;            if(arr[i].y > maxtime)  maxtime = arr[i].y;        }        for(i=1; i<=maxtime; i++){            pre[i] = i;        }        sort(arr, arr+n);        memset(vis, 0, sizeof(vis));        int s = 0;        for(i=0; i<n; i++){            int d = find(arr[i].y);            if(d>0){                s += arr[i].x;                pre[d] = d-1;            }        }        printf("%d\n", s);    }    return 0;}