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.
輸入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.輸出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.樣例輸入4 50 2 10 1 20 2 30 17 20 1 2 1 10 3 100 2 8 2 5 20 50 10樣例輸出80185解題報告：這道題的意思是，每件商品賣出需花費一天時間，只要在截止日期之前（包含）賣出就行。將價值降序，截止日期升序這種方法是不行的。
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天是否有賣商品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--){ //在截止日期之前賣出就行 if(visited[j]) continue; visited[j]=1; sum+=a[i].p; break; } } printf("%d/n",sum); } return 0;}
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