Doing Homework again【HDU - 1789】

最新推荐文章于 2020-09-07 17:28:41 发布

原创最新推荐文章于 2020-09-07 17:28:41 发布 · 194 阅读

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本文介绍了一个典型的贪心算法问题，即如何合理安排家庭作业的完成顺序以最小化因延迟提交而导致的扣分。通过遍历每个作业的截止日期并优先处理高分值任务来解决此问题。

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

题解：

典型贪心。给了所有作业的截止时间和惩罚的扣分分值，求最小的扣分。根据题意，需优先完成分值大的作业。

以截止时间开始向前遍历，依次遍历，最后把无法完成的作业的分值加起来，输出。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int a,b;
}edg[1010];
bool cmp(node x,node y)
{
    if(x.b==y.b)
        return x.a<y.a;
    return x.b>y.b;
}
int main()
{
    int t,n,m,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%d",&edg[i].a);
        for(i=0;i<n;i++)
            scanf("%d",&edg[i].b);
        sort(edg,edg+n,cmp);
        int book[1010]={0};
        int ans=0;
        for(i=0;i<n;i++)
        {
            for(j=edg[i].a;j>=1;j--)
            {
                if(!book[j])
                {
                    book[j]=1;
                    break;
                }
            }
            if(j<=0)
                ans+=edg[i].b;
        }
        printf("%d\n",ans);
    }
}