杭电1789—Doing Homework again

最新推荐文章于 2020-05-30 07:04:26 发布

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最新推荐文章于 2020-05-30 07:04:26 发布

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本文介绍了一个通过合理安排作业顺序来最小化逾期扣分的算法。该算法首先按照扣分多少和截止日期进行排序，然后尝试将每项作业安排在最接近其截止日期且未被占用的时间段内。

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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

10 5 1

1 3 1

6 2 3

1 4 6 4 2 4 3

3 2 1 7 6 5 4

/*
    此题大致思路，既然要计算最少扣多少分，就眼前利益
    考虑必然要先把超过最后时间扣分最多的作业先安排了
    如果扣分一样多的话，那必然要把时间比较紧的先安排
    了.所以先按扣分的高低，由高向低排序，如果两门课
    扣分相同就按他们的结束时间有低向高排序!然后选择即可！ 
*/
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
	int day,score;
}q[1010];
int flag[1010];
bool cmp(node x,node y)
{
	if(x.score!=y.score) return x.score>y.score;
	else return x.day<y.day;
}
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int time;
		int ans=0;
		memset(flag,0,sizeof(flag));
		cin>>time;
		memset(q,0,sizeof(q));
		for(int i=0;i<time;i++) cin>>q[i].day;
		for(int i=0;i<time;i++) cin>>q[i].score;
		sort(q,q+time,cmp);
		for(int i=0;i<time;i++)
		{
			int j;
			for(j=q[i].day;j>0;j--)
			{
				if(!flag[j])
				{
					flag[j]=1;
					break;
				}
			}
			if(j==0)
			{
				ans+=q[i].score;
			} 
		}
		cout<<ans<<endl;
	}
	return 0;
}