Doing Homework again

Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test. And now we assume that doing everyone homework always takes one day. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.InputThe input contains several test cases. The first line of the input is a single integer T that is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=1000) which indicate the number of homework.. Then 2 lines follow. The first line contains N integers that indicate the deadlines of the subjects, and the next line contains N integers that indicate the reduced scores.
OutputFor each test case, you should output the smallest total reduced score, one line per test case.
Sample Input

3
3
3 3 3
10 5 1
3
1 3 1
6 2 3
7
1 4 6 4 2 4 3
3 2 1 7 6 5 4

Sample Output

0
3
5

非常体现贪心的思想，重点就是贪的策略和时间的转换，扣分高的优先完成，扣分相同而时间少的优先完成。然后看是否能在当前作业的deadline之前完成，而将时间从后向前检查，目的就是尽量在后面的天数
完成当前作业。这点很重要！

#include<cstdio> 
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct node{
    int time,grade;
    bool operator<(const node & i)const{
        if(grade==i.grade) return time<i.time;
        return grade>i.grade;
    }
}cla[1005];

int cases;
int n;
int vis[1005];

int main()
{   cin>>cases;
    while(cases--){
        cin>>n;
        for(int i=0;i<n;i++) scanf("%d",&cla[i].time);
        for(int i=0;i<n;i++) scanf("%d",&cla[i].grade);
        
        memset(vis,0,sizeof(vis));
        sort(cla,cla+n);
        int ans=0;
        for(int i=0;i<n;i++){
            int temp=cla[i].time;
            bool flag=false;
            while(temp){
                if(!vis[temp]) {vis[temp]=true; flag=true; break; }
                temp--; 
            }
            if(!flag) ans=ans+cla[i].grade;
        }
        cout<<ans<<endl;
    }
    return 0;
}