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：
每行第一个数字是商品件数，后面每两个数字未一个商品的价值，如case1:
4 50 2 10 1 20 2 30 1
代表 4 件商品
数据分别为：
价值 售卖时间
50 2
10 1
20 2
30 1
不可对时间贪心，因为时间靠后的商品如果价值高是可以拿到前面来销售的
因此对价格进行贪心:
如果当前最高价格在时间没有商品供销售，则销售它，如果已经有更高价格的商品在这个时间销售，则把它往前挪，把它放到时间靠前的时间销售（因为是按价格优先的，所以当前它的销售优先级最高咯）。
代码：
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<list>
#include<iterator>
#include<stack>
#include <queue>
#include <cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define e 2.718281828459
#define INF 0x7fffffff
#pragma warning(disable:4996)
#define sf scanf
#define pf printf
const double pi = acos(-1.0);
const int MAX = 10005;
//#define eps 1e-9;
struct Node {
int pro;
int ti;
friend bool operator < (const Node& a,const Node& b) {
if (a.pro == b.pro)
return a.ti < b.ti;
else
return a.pro < b.pro;
}
};
int main(void) {
int n;
bool vis[MAX];
Node temp;
priority_queue<Node> da;
while (sf("%d", &n)!=EOF) {
ull ans = 0;
memset(vis, 0, sizeof vis);
for (int i = 0; i < n; i++) {
sf("%d %d", &temp.pro, &temp.ti);
da.push(temp);
}
for (int i = 0; i < n; i++) {
temp = da.top();
if (!vis[temp.ti]) {//这个时间还没有卖过
ans += temp.pro;
vis[temp.ti] = true;
da.pop();
}
else {
while (--temp.ti) {//往前卖
if (!vis[temp.ti] && temp.ti >= 1) {//前面这个时间还没有卖过
ans += temp.pro;
vis[temp.ti] = true;
break;//记得跳出，不然相当于多出商品，我就是在这里WA了半天(temp.ti会一直--，直到0，这是很糟糕的)
}
}
da.pop();//不论买没卖出去，都要删了
}
}
pf("%lld\n", ans);
while (!da.empty())
da.pop();
}
return 0;
}