Doing Homework again 【贪心】

最新推荐文章于 2020-12-07 23:13:57 发布

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本文探讨了如何通过合理安排作业顺序来最小化因逾期提交而导致的扣分问题。采用分数优先级排序策略，确保高分值作业尽可能在截止日期前完成。通过对多个测试案例的分析，展示了该算法的有效性和实用性。

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O - Doing Homework again

HDU - 1789

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

思路：分数优先由大到小排。分数大的那个截止日期，一定要是由它用掉，如果当前的时间用过，就向前找，看有没有没有用的，，

代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<vector>
#define inf 0x3f3f3f
#define mod 100000
#define M 1000+10
using namespace std;
struct data
{
	int day,score;
};
data a[M];
int cmp(data a,data b)
{
	return a.score>b.score;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		int days=0;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i].day);
			days=max(days,a[i].day);		
		}
		int v[10000]={0};
		for(int i=0;i<n;i++)
		scanf("%d",&a[i].score);
		sort(a,a+n,cmp);
		int sum=0;
		for(int i=0;i<n;i++)
		{
			if(v[a[i].day]==0)
			v[a[i].day]=1;
			else 
			{		int j;
				for( j=a[i].day-1;j>=1;j--)
				if(v[j]==0) 
				{
					v[j]=1;
					break;
				}
				if(j==0) sum+=a[i].score;
			}
		}
		printf("%d\n",sum);
	}
	return 0;
 }