Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

思路：

既然要让被扣掉的分数最少，那么必然是对分数高的作业优先安排。注意题中有一个不是很明显的条件可以支持这一点：完成每份作业都需要一天。这样就避免了优先完成一份分数高的作业而导致n（n>1）份作业没有完成，而且这n份作业分数和比一份分数高的作业还要大的情况。

方法：

对于所有的作业，按照分数从高到低排序，分数相同时，截止时间小的排在前面。另外初始化一个大小为n的数组，用来保存某一天是否已经被占用。然后开始贪心，对于每份作业，看从当天到当前之前的时间里面，有没有空余，如果有空余，安排到空余且天数最大的一天；如果没有空余，这份作业只能被扣分。注意这边天数从1开始，而代码中习惯数组从0开始，天数减1或者数组加1都可以。

参考代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node{
    int deadline;
    int score;
};
bool cmp(node a, node b){
    if(a.score==b.score){
    	return a.deadline<b.deadline;
    }
    return a.score>b.score;
}
node homework[1002];
int used[1002];
int main(){
	int t, n, ans;
	scanf("%d", &t);
	while(t--){
		scanf("%d", &n);
		for(int i=0;i<n;i++){
			scanf("%d", &homework[i].deadline);
		}
		for(int i=0;i<n;i++){
			scanf("%d", &homework[i].score);
		}
		sort(homework, homework+n, cmp);
		for(int i=0;i<n;i++){
			used[i] = 0;
		}
		ans = 0;
		for(int i=0;i<n;i++){
			bool flag = false;
			for(int j=homework[i].deadline-1;j>=0;j--){
				if(used[j]==0){
					used[j] = 1;
					flag = true;
					break;
				}
			}
			if(!flag){
				ans += homework[i].score;
			}
		}
		printf("%d\n", ans);
	}
}