Doing Homework again

题目：

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

 

大意：

有几门作业需要在规定的天数内完成，每天只能且只能完成一门作业；

如果未在规定时间内完成，就需要扣掉相应分数；

问，最少扣多少分？

 

思路：

运用贪心思想，先对每门课超时所扣分进行从大到小排列；

然后，从头到尾遍历，规定每门课都尽量在截止日期当天完成，如果当天时间已经被占用，则往前一天推，看 在截至日期以前是否能够完成，如果不能完成，则所扣总分加上这门课要扣的分数；

 

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int x,y;
}e[100861];
bool cmp(node a,node b)
{
    return a.y>b.y;
}
int book[100861];
int main()
{
    int m;
    scanf("%d",&m);
    while(m--)
    {
        int n,i,sum=0;
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%d",&e[i].x);
        for(i=0;i<n;i++)
            scanf("%d",&e[i].y);
        sort(e,e+n,cmp);
        memset(book,0,sizeof(book));
        for(i=0;i<n;i++)
        {
            int flag=0;
            for(int j=e[i].x;j>0;j--)
            {
                if(book[j]==0)
                {
                    book[j]=1;
                    flag=1;
                    break;
                }
            }
            if(flag==0)
                sum=sum+e[i].y;
        }
        printf("%d\n",sum);
    }
    return 0;
}