编程竞赛题目：做作业的贪心算法-优快云博客

这是一个编程竞赛的题目，名为'Doing Homework again'，涉及到使用贪心算法解决问题。比赛对时间限制和内存限制有特定要求，并且已经收到了一定数量的提交和接受解决方案。

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Doing Homework again

Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 4 Accepted Submission(s) : 3

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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 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

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct node{
	int t,s;
}f[1005];
int vis[1005];
int cmp(node a,node b){
	return a.s >b.s ;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--){
		int n,sum=0;
		memset(vis,0,sizeof(vis));
		scanf("%d",&n);
		for(int i=0;i<n;i++)
			scanf("%d",&f[i].t );
		for(int i=0;i<n;i++)
			scanf("%d",&f[i].s );
		sort(f,f+n,cmp);//分值从大到小排 
		for(int i=0;i<n;i++){
			if(vis[f[i].t]==0) //让其在当天完成
			vis[f[i].t]=1;
			else{//如果这一天被占用了， 
				int flag=0;
				for(int j=f[i].t ;j>=1;j--){//依次往前看有没有空的一天， 
					if(vis[j]==0){//因为目前作业分值最大，优先考虑 
						vis[j]=1;
						flag=1;
						break;
					}
				}
				if(flag==0)//没有空闲，只能放弃 
				sum+=f[i].s;
			}
		}
		printf("%d\n",sum);
	}
	return 0;
}