Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15893    Accepted Submission(s): 9247

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


这题与 HDU - 1051 题 思路差不多，先将数组按扣分多少从大到小排序，扣分相同时将时间按从小到大排，然后循环求解。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct dead{
	int day,red;
};

bool cmp(dead a, dead b)
{
	if (a.red!=b.red)
		return a.red > b.red;
	return a.day < b.day;
}
int wri[1005];
int main()
{
	int T,N;
	scanf("%d", &T);
	struct dead line[1005];
	while (T--)
	{
		scanf("%d", &N);
		for (int i = 0; i < N; i++)
		{
			scanf("%d", &line[i].day);
		}
		for (int i = 0; i < N; i++)
		{
			scanf("%d", &line[i].red);
		}
		sort(line, line + N, cmp);
		int sum = 0,j=0;
		memset(wri, 0, sizeof(wri));
		for (int i = 0; i < N; i++)
		{
			j = line[i].day;
			while (j)
			{
				if (!wri[j])//判断这天是否写过作业
				{
					wri[j] = 1;
					break;
				}
				j--;
			}
			if (j == 0)
				sum += line[i].red;
		}
		printf("%d\n", sum);
	}
	return 0;
}