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

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

struct Node
{
    int time,fen;
} node[1005];

int cmp(struct Node a,struct Node b)
{
    if(a.fen!=b.fen)
        return a.fen > b.fen;//扣分越多的越靠前
    return a.time < b.time;//扣分相同的時候,deadline越早的越靠前
}

int  visit[2010];//如果當天沒用過,值為0；否則為1

int main()
{
    int m;
    cin >> m;
    while(m--)
    {
        int n,i,j,ans = 0;
        memset(visit,0,sizeof(visit));
        cin >> n;
        for(i = 0; i<n; i++)
        {
            cin >> node[i].time;
        }
        for(i = 0; i<n; i++)
        {
            cin >> node[i].fen;
        }
        sort(node,node+n,cmp);
        int sum = 0;
        for(i=0; i<n; i++)
        {
            j=node[i].time;// 從截止時間開始往前推,如果有一天沒用過,這一天就做這一門課,這門課不扣分
            while(j)
            {
                if(!visit[j])
                {
                    visit[j]=1;
                    break;
                }
                j--;
            }
            if(j==0)//如果j=0,表明從time往前的每一天都被佔用了,這門課完不成
                ans+=node[i].fen;
        }
        cout << ans << endl;
    }

    return 0;
}