Doing Homework again

最新推荐文章于 2020-04-06 11:28:10 发布

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最新推荐文章于 2020-04-06 11:28:10 发布

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本文链接：https://blog.youkuaiyun.com/u011471344/article/details/9732769

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本文介绍了一个基于截止日期和惩罚分数的作业调度问题。通过合理的排序算法，先按惩罚分数从大到小排序，分数相同时按截止日期从小到大排序，从而实现作业调度的最优化，最小化总惩罚分数。

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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
   
   #include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;
int m,n;
struct tv
{
    int a;
    int b;
    double num;
} t[1010];
int flag[1010];
bool cmp(const tv a, const tv b)
{
    if(a.b != b.b)  return a.b > b.b;
    return a.a < b.a;
}
int main()
{
    scanf("%d",&m);
    while(m--)
    {
        scanf("%d",&n);
        memset(flag, 0, sizeof(flag));
        int sum1=0,i,j;
        for(i=0; i<n; i++)
            scanf("%d",&t[i].a);
        for(i=0; i<n; i++)
        {
            scanf("%d",&t[i].b);
        }
        sort(t,t+n,cmp);
        for( i = 0; i < n; i++)
        {

            for(j = t[i].a; j > 0; j--)
            {
                if(flag[j] == 0)
                {
                    flag[j] = 1;
                    break;
                }
            }
            if(j == 0)
                sum1+= t[i].b;
        }
        printf("%d",sum1);
            printf("\n");
    }
}

  
  
  
  
   
    
  
  
  
  
   
    
  
  
  
  
   
    
  
  
  
  
   
    
  
  
  
  
   
    
  
  
  
  
   
    
  
  
  
  
   
   排序原理  ： 以分数排序   ，若分数相同，则依靠限制天数来排序。
  
  
  
  
   
    
  
  
  
  
   
    
  
  
  
  
   
   先拍分数最大的       尽量排在当天    如果不能排在当天   则向前推   知道零  则表示不能输入了    完成  输出答案