Doing Homework again

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.

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

题意：有几门课程的作业，每门都有限制时间，如果不在截止时间之前做好，就会扣掉那门课程作业相应的分，怎样安排使得扣分最少；

思路：只有把分值高的的作业在截止时间前做好就可以，因此先对分值做从大到小的排序；

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
	int a,b;
}s[1010];
bool cmp(node x,node y)
{
	if(x.b!=y.b)
    	return x.b>y.b;//作业分从大到小排序 
	else
	    return x.a<y.a;//当分值相同时，把时间从小到大排序 
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int i,j,n;
		int sum=0,s1[1010]={0};
		scanf("%d",&n);
		for(i=0;i<n;i++)
		   scanf("%d",&s[i].a);
		for(i=0;i<n;i++)
	     	scanf("%d",&s[i].b);
		sort(s,s+n,cmp);
		for(i=0;i<n;i++)
		{
			for(j=s[i].a;j>0;j--) 
			{
				if(s1[j]==0)//在截止时间之前，有没被标记的，就代表着那天是空闲的 
				{
					s1[j]=1;//把这天安排作业，标记为1； 
					break;
				}
			}
			if(j==0)//在开始到截止时间为止都被标记，说明没有空闲时间，就要扣分； 
			sum+=s[i].b;
		}
		printf("%d\n",sum);
	}
}