Doing Homework again

最新推荐文章于 2020-12-07 23:13:57 发布

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本文介绍了一个关于如何使用贪心算法来最小化因延迟交作业而被扣分的问题。通过对作业截止日期和扣分进行排序，算法确保了学生可以在有限时间内完成作业，同时尽可能减少总扣分。

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

Author

lcy

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

题意：

本题考察的的是贪心问题，要想减得分数最少，先找做分数最多的（因为无论哪一门都是一天做完），若分数相同，则先做时间最早的。这样才能损失的最少。

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
struct course{
	int time;
	int score;
}a[100000];
int cmp(course b,course c)
{
	if(b.score!=c.score)
	return b.score>c.score;
	return b.time<c.time;
}
int f[10000]={0};
int main(){
	int n,m,i,j,k,sum;
	scanf("%d",&n);
	while(n--)
	{
		sum=0;
		memset(f,0,sizeof(f));
		scanf("%d",&m);
		for(i=0;i<m;i++)
		scanf("%d",&a[i].time);
		for(j=0;j<m;j++)
		scanf("%d",&a[j].score);
	    sort(a,a+m,cmp);
	    for(i=0;i<m;i++)
	    {
	    	for(j=a[i].time;j>0;j--)  //   从截止日期里找没有占用掉的时间。
	    	{
	    		if(f[j]==0)
	    		{
	    			f[j]=1;
	    			break;
	    		}
	    	}
	    	if(j==0)                         //如果找不到，则加到罚分里面。
	    	sum+=a[i].score;
	    }
	    printf("%d\n",sum);
	}
	return 0;
}