HDU - 1789 Doing Homework again 贪心

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： HDU 文章标签： ACM 贪心

本文链接：https://blog.youkuaiyun.com/leo6033/article/details/79156955

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本文介绍了一个通过合理安排作业顺序来最小化因逾期而扣分的算法。该算法首先根据扣分大小及截止日期对作业进行排序，然后通过循环遍历的方式确定每个作业的最佳完成时间，最终计算出总的扣分情况。

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Doing Homework again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15893 Accepted Submission(s): 9247

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

这题与 HDU - 1051 题思路差不多，先将数组按扣分多少从大到小排序，扣分相同时将时间按从小到大排，然后循环求解。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct dead{
	int day,red;
};

bool cmp(dead a, dead b)
{
	if (a.red!=b.red)
		return a.red > b.red;
	return a.day < b.day;
}
int wri[1005];
int main()
{
	int T,N;
	scanf("%d", &T);
	struct dead line[1005];
	while (T--)
	{
		scanf("%d", &N);
		for (int i = 0; i < N; i++)
		{
			scanf("%d", &line[i].day);
		}
		for (int i = 0; i < N; i++)
		{
			scanf("%d", &line[i].red);
		}
		sort(line, line + N, cmp);
		int sum = 0,j=0;
		memset(wri, 0, sizeof(wri));
		for (int i = 0; i < N; i++)
		{
			j = line[i].day;
			while (j)
			{
				if (!wri[j])//判断这天是否写过作业
				{
					wri[j] = 1;
					break;
				}
				j--;
			}
			if (j == 0)
				sum += line[i].red;
		}
		printf("%d\n", sum);
	}
	return 0;
}