acm之贪心算法题目5

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本文介绍了一个关于ACM竞赛后Ignatius面临的作业问题。每个作业都有截止日期，逾期提交会导致期末成绩扣分。目标是确定完成作业的顺序，以最大限度地减少总扣分。输入包含多个测试用例，每个用例包含作业数量、各科截止日期及相应扣分。输出是最小的总扣分。样例输入和输出展示了不同场景下的计算结果。

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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

代码：

#include<stdio.h>
#include<string.h>
#define max 1005
struct he
{
   int time;
   int score;
};
bool f[max];
int main()
{
    int t,n,i,min,j;
    he student[max],temp;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        scanf("%d",&student[i].time);
        for(i=0;i<n;i++)
           scanf("%d",&student[i].score);
        min=0;
        memset(f,0,sizeof(f));
        for(i=0;i<n;i++)
           for(j=i+1;j<n;j++)
           {
                if(student[j].score>student[i].score)               
                { 
                   temp=student[j];
                   student[j]=student[i];
                }
                else if(student[j].score==student[i].score  
                && student[j].time<student[i].time)
               {  temp=student[j];
                  student[j]=student[i];
                  student[i]=temp; }
            }
         for(i=0;i<n;i++)
         {
             for(j=student[i].time;j>0;j--)
             {
                    if(f[j]==false)
                    {
                         f[j]=true;
                         break;
                    }
              }
              if(j==0)
                    min+=student[i].score;
         }
              printf("%d\n",min);
    }
    return 0;
}