Doing Homework again 贪心算法

最新推荐文章于 2020-12-07 23:13:57 发布

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本文介绍了一个简单的贪心算法，用于帮助学生合理安排多门课程的作业完成顺序，以最小化因逾期提交而导致的成绩扣分。算法通过优先处理扣分较高的作业来实现这一目标。

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.

输入：

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.

输出：

For each test case, you should output the smallest total reduced score, one line per test case.

样例输入：

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

样例输出：

0
3
5

很简单的一个贪心算法，从价值最大的作业开始安排

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int v[10000];
struct work
{
    int va;
    int day;
};
bool cmp(work a,work b)
{
    return a.va>b.va;
}
int so(int a)                         ///对不同的天数进行访问
{
    if(a==0)                         ///当天数为零时，无法安排，只能舍弃
    {
        return 1;
    }
    else if(v[a]==0)                ///当找到某一天空闲时，将作业安排上，并标记访问
    {
        v[a]=1;
        return 0;
    }
    else if(v[a]==1)              ///如果当前天数被访问，那么继续向前找
    {
        so(a-1);
    }
}
int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int m;
        scanf("%d",&m);
        work ss[m];
        for(int i=0;i<=m;i++)
        {
            v[i]=0;
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d",&ss[i].day);
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d",&ss[i].va);
        }
        sort(ss,ss+m,cmp);
        int sum=0;
        for(int j=0;j<m;j++)
        {
            if(so(ss[j].day))
            {
                sum=sum+ss[j].va;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}