HDOJ 1789 Doing Homework again

最新推荐文章于 2018-10-07 16:21:00 发布

原创最新推荐文章于 2018-10-07 16:21:00 发布 · 288 阅读

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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。该算法首先将作业按分数降序排列，然后尝试将每个作业安排在截止日期当天完成。若该天已满，则向前寻找最早可安排的时间，若所有时间均不可用则计入扣分。

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Doing Homework again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11320 Accepted Submission(s): 6649

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

Author

lcy

题解：

开一个代表天数的数组。

把题目按分数降序排列，然后把可以刚好压着死线的天数标记。如果那一天被标记了，那么就往前推，如果前面所有的天数都被标记了，那么说了这项作业做不完，要被扣分。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct cl
{
	int day,score;
}a[1111];
int d[1111];
bool cmp(cl a,cl b)
{
	if(a.score==b.score)
		return a.day>b.day;
	return a.score>b.score;
}
int main()
{
	int T;
	int i,j,n;
	scanf("%d",&T);
	while(T--)
	{
		memset(d,0,sizeof(d));
		scanf("%d",&n);
		for(i=0;i<n;i++)
			scanf("%d",&a[i].day);
		for(i=0;i<n;i++)
			scanf("%d",&a[i].score);
		sort(a,a+n,cmp);
		int flag=1;
		int t=0;
		for(i=0;i<n;i++)
		{
			if(d[a[i].day]==0)
				d[a[i].day]=1;
			else
			{
				for(j=a[i].day;j>0;j--)
				{
					if(d[j]==0)
					{
						d[j]=1;
						flag=0;
						break;
					}
				}
				if(flag)
					t+=a[i].score;
				flag=1;
			}
		
		}
		printf("%d\n",t);
	}
	return 0;
}