HDU 1789 Doing Homework again(馋)

博客介绍了如何通过合理安排作业顺序,以最小化因提交超时而引起的分数损失,适用于ACM/ICPC竞赛场景。

意甲冠军  参加大ACM竞争是非常回落乔布斯  每一个工作都有截止日期   未完成必要的期限结束的期限内扣除相应的积分   求点扣除的最低数量

把全部作业按扣分大小从大到小排序  然后就贪阿  能完毕前面的就完毕前面的  实在不能的就扣分吧~

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1005;
int dli[N], red[N], k[N], cas, ans, n;
bool vis[N];

bool cmp (int i, int j)
{
    return red[i] > red[j];
}

int main()
{
    scanf ("%d", &cas);
    while (cas--)
    {
        ans = 0;
        memset (vis, 0, sizeof (vis));
        scanf ("%d", &n);
        for (int i = 1; i <= n; ++i)
            scanf ("%d", &dli[i]), k[i] = i;
        for (int j = 1; j <= n; ++j)
            scanf ("%d", &red[j]);
            
        sort (k + 1, k + n + 1, cmp);
        for (int i = 1, j; i <= n; ++i)
        {
            for (j = dli[k[i]]; j >= 1; --j)
                if (!vis[j])
                {
                    vis[j] = 1;
                    break;
                }
            if (j == 0) ans += red[k[i]];
        }
        printf ("%d\n", ans);
    }
    return 0;
}<span style="font-family:Comic Sans MS;">
</span>

Doing Homework again


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
 



版权声明:本文博主原创文章。博客,未经同意不得转载。

转载于:https://www.cnblogs.com/gcczhongduan/p/4912591.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值