Doing Homework again

最新推荐文章于 2020-04-06 11:28:10 发布

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本文介绍了一个通过合理安排作业顺序来最小化逾期扣分的算法。该算法首先对作业按扣分大小进行排序，然后尝试在每个作业的截止日期前完成。若某项作业无法在期限内完成，则计入扣分。最终输出的是最小化的总扣分。

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E - Doing Homework again

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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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

这个题是多组输入的多组输入。这个题的大意就是：总共有N科作业，每科都有一定的期限，每科都有没有完成情况下的扣分，求在相应的期限里最少的扣分。

#include <iostream>
#include <string.h>
using namespace std;
bool tt[1050];
struct node
{
    int date,grade;
} dd[1050],temp;
int main()
{
    int T;
    while(scanf("%d",&T)!=EOF)
    {
        while(T--)
        {
            int n,i,j;
            scanf("%d",&n);
            memset(tt,0,sizeof(tt));
            for(i=0; i<n; i++)
                scanf("%d",&dd[i].date);
            for(i=0; i<n; i++)
                scanf("%d",&dd[i].grade);
            for(i=0; i<n-1; i++)                   //按扣分的多少排序，扣分最多的在最前边，扣分相同的情况下，期限最短的在前边
                for(j=0; j<n-1-i; j++)
                {
                    if(dd[j].grade<dd[j+1].grade)
                    {
                        temp=dd[j];
                        dd[j]=dd[j+1];
                        dd[j+1]=temp;
                    }
                    else if(dd[j].grade==dd[j+1].grade)
                    {
                        if(dd[j].date>dd[j+1].date)
                        {
                            temp=dd[j];
                            dd[j]=dd[j+1];
                            dd[j+1]=temp;
                        }
                    }
                }
            int sum=0;
            for(i=0; i<n; i++)                     //这几句你按实际操作手写走一遍就明白什么意思了，我说也说不明白
            {
                int flag=0;
                for(j=dd[i].date;j>0; j--)
                {
                    if(tt[j]==0)
                    {
                        tt[j]=1;
                        flag=1;
                        break;
                    }
                }
                if(!flag)
                    sum+=dd[i].grade;
            }
            printf("%d\n",sum);
        }
    }
    return 0;
}