HDU - 1789 Doing Homework again

最新推荐文章于 2019-03-12 18:05:39 发布

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最新推荐文章于 2019-03-12 18:05:39 发布

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分类专栏：贪心文章标签： HDU - 1789 Doing Homework again

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本文介绍了一个关于作业排序的问题，通过贪心算法实现作业最优排序，以最小化因逾期而产生的分数损失。

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Doing Homework again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14230 Accepted Submission(s): 8269

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

题意：

老师布置了n道家庭作业，每个作业有截至日期和分数，如果到截止日期还没有完成，就要扣掉相应的分数。每个作业耗时1天。

题解：

直接枚举天数。按照贪心思想，先将作业排序，优先将分数从大到小排序，如果分数相同则将deadline从小到大排序(多余的一逼)。两层循环，第一层枚举作业，第二层从当前作业的deadline往前枚举天数，如果这一天没有安排（即没被标记），就可以将这个作业安排在这一天，如果前面几天都有安排，则将score加到sum里。代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int deadline,score;
}p[1005];
bool vis[2000];
bool cmp(node a,node b)
{
    return a.score==b.score? a.deadline<b.deadline:a.score>b.score;
}
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&p[i].deadline);
        for(int i=0;i<n;i++)
            scanf("%d",&p[i].score);
        sort(p,p+n,cmp);
        memset(vis,0,sizeof(vis));
        int res=0,flag;
        for(int i=0;i<n;i++)
        {
            flag=0;
            for(int j=p[i].deadline;j>0;j--)
            {
                if(!vis[j])
                {
                    vis[j]=1;
                    flag=1;
                    break;
                }
            }
            if(!flag)
                res+=p[i].score;
        }
        printf("%d\n",res);
    }
    return 0;
}