hdu 1789 Doing Homework again

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
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    这道是一道简单的贪心问题；
   
   
   
   
   
   
   
   

    
    
    
    按分数排序，完成日期优先安排在截止日期；
   
   
   
   
   
   
   
   

    
    
    
    若已被占，就向前移；
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    
#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<utility>
using namespace std;
struct Work
{
    int d;
    int p;
};
bool cmp(const Work &a,const Work &b)
{
    return a.p>b.p;
}
int main()
{
    int T;
    scanf(" %d",&T);
    while(T--)
    {
        int N,ans=0;
        bool flag[10000]={0};
        Work work[1005];
        scanf(" %d",&N);
        int a,b;
        for(int i=0;i<N;++i)
        {
            scanf(" %d",&work[i].d);
        }
        for(int i=0;i<N;++i)
        {
            scanf(" %d",&work[i].p);
        }
        sort(work,work+N,cmp);
        for(int i=0;i<N;++i)
        {
            int j;
            for(j=work[i].d;j>=1;--j)
            {
                if(flag[j]==0)
                {
                    flag[j]=1;
                    break;
                }
            }
            if(j==0) ans+=work[i].p;
        }
        printf("%d\n",ans);
    }
    return 0;
}