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.
输入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.输出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.样例输入4 50 2 10 1 20 2 30 17 20 1 2 1 10 3 100 2 8 2 5 20 50 10样例输出80185解题报告：这道题的意思是，每件商品卖出需花费一天时间，只要在截止日期之前（包含）卖出就行。将价值降序，截止日期升序这种方法是不行的。
code：
#include<iostream>#include<stdio.h>#include<queue>#include<vector>#include<stack>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const int maxn=10005;int visited[maxn]; //代表第i天是否有卖商品struct node{ int p,d;};bool cmp(node a,node b){ return a.p>b.p;}int main(){ // freopen("input.txt","r",stdin); int n; while(~scanf("%d",&n)){ memset(visited,0,sizeof(visited)); node a[maxn]; //存商品 for(int i=0;i<n;i++){ scanf("%d%d",&a[i].p,&a[i].d); } sort(a,a+n,cmp); int sum=0; for(int i=0;i<n;i++){ for(int j=a[i].d;j>=1;j--){ //在截止日期之前卖出就行 if(visited[j]) continue; visited[j]=1; sum+=a[i].p; break; } } printf("%d/n",sum); } return 0;}
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