题目链接：

啊哈哈，点我点我

思路：

这道题是简单的贪心。。先按分数从大到小排序，然后将这个分数的截止日期从后向前扫描，如果碰到没有被标记的则这一天可以做这个作业。。。

题目：

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

Author

lcy

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

Recommend

lcy

代码为：

#include<cstdio>
#include<map>
#include<iostream>
#include<algorithm>
using namespace std;

map<int,bool>mp;

struct homework
{
    int deadline,score;
}home[1000+10];

bool cmp(homework a,homework b)
{
   if(a.score==b.score)
        return a.deadline<b.deadline;
   else
        return a.score>b.score;
}

int main()
{
    int n,ans,l,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        ans=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&home[i].deadline);
            mp[i]=false;
        }
        for(int i=1;i<=n;i++)
            scanf("%d",&home[i].score);
        sort(home+1,home+1+n,cmp);
        for(int i=1;i<=n;i++)
        {
            for(l=home[i].deadline;l>=1;l--)
            {
                if(!mp[l])
                {
                    mp[l]=true;
                    break;
                }
            }
            if(l<1)  ans=ans+home[i].score;
        }
        printf("%d\n",ans);
    }
    return 0;
}

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