hdoj 1789 Doing Homework again

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8638    Accepted Submission(s):
5090

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

先对分数从大到小排序 然后设出一个数组vis[]来按照分数从高到低记录必须占用的日期如果有空闲的日期

则标记为1 若没有空闲的日期则舍弃这个分数

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct record
{
	int day;
	int score;
}num[1100];
bool cmp(record a,record b)
{
	if(a.score!=b.score)
	return a.score>b.score;
	else
	return a.day<b.day;
}
int main()
{
	int n,m,j,i,t,l,sum;
	int vis[1100];//用来储存必须要占用的日期
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		memset(vis,0,sizeof(vis));
		for(i=0;i<n;i++)
			scanf("%d",&num[i].day);
		for(i=0;i<n;i++)
		    scanf("%d",&num[i].score);
		sort(num,num+n,cmp);
		sum=0;
		for(i=0;i<n;i++)
		{
			for(j=num[i].day;j>=1;j--)
			{
				if(vis[j]==0)
				{
					vis[j]=1;
					break;
				}
			}
			if(j==0)
			sum+=num[i].score;
		}
		printf("%d\n",sum);
	}
	return 0;
}