hdu-1789-Doing Homework again

/*
Doing Homework again
Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4568    Accepted Submission(s): 2675


Problem Description
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. 
 

Input
The 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.

 

Output
For 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
 

Author
lcy
 

Source
2007省赛集训队练习赛（10）_以此感谢DOOMIII 
 

Recommend
lcy

本题和前面一道贪心是一样的
*/

#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<string>
using namespace std;
#define maxn 1010
struct DL
{
    int a;
    int b;
    bool operator <(const DL &d)const
    {
        return d.b<b;
    }
} dl[maxn];
bool v[maxn];
int main()
{
    int t,i,n,sum,j;
    scanf("%d",&t);
    while (t--)
    {
        memset(v,0,sizeof(v));
        scanf("%d",&n);
        for(i=0; i<n; i++)
            scanf("%d",&dl[i].a);
        for(i=0; i<n; i++)
            scanf("%d",&dl[i].b);
        sort(dl,dl+n);
        sum=0;
        for(i=0;i<n;i++)
        {
            for(j=dl[i].a;j>0;j--)
            {
                if(!v[j])
                {
                    v[j]=1;
                    break;
                }

            }
            if(j==0)
            {
                sum+=dl[i].b;
            }
        }
        printf("%d
",sum);
    }
    return 0;
}