HDU1789Doing Homework again
来源:互联网 发布:易语言2017盗号源码 编辑:程序博客网 时间:2024/06/01 17:21
Doing Homework again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15250 Accepted Submission(s): 8903
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.
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
333 3 310 5 131 3 16 2 371 4 6 4 2 4 33 2 1 7 6 5 4
Sample Output
035
Author
lcy
Source
2007省赛集训队练习赛(10)_以此感谢DOOMIII
#include<iostream>#include<algorithm>#include<cstring>using namespace std;struct stu{int end,fen;}arr[1010];bool cmp(stu x,stu y){if(x.fen==y.fen)return x.end<y.end;return x.fen>y.fen;}int main(){ int T,a[1010];cin>>T;while(T--){int N,ans=0,j;cin>>N;memset(a,0,sizeof(a));for(int i=0;i<N;i++)cin>>arr[i].end;for(int i=0;i<N;i++)cin>>arr[i].fen;sort(arr,arr+N,cmp);for(int i=0;i<N;i++){for(int j=arr[i].end;j>=1;j--){if(a[j]==0){a[j]=1;break;}if(j==1)ans+=arr[i].fen;}}cout<<ans<<endl;} }
参考别人的博客题解
先按照扣分从大到小排序,分数相同则按照截止日期从小到大排序。。
然后按顺序,从截止日期开始往前找没有占用掉的时间。
如果找不到了,则加到罚分里面
单独设置一个数组保存这一天是否完成作业。。。。如果没有安排,就在这一天做。
阅读全文
0 0
- HDU1789Doing Homework again
- hdu1789doing homework again【贪心】
- hdu1789Doing Homework again
- HDU1789Doing Homework again
- hdu1789Doing Homework again【贪心算法】
- hdu1789Doing Homework again(贪心)
- HDU1789Doing Homework again(贪心)
- hdu1789Doing Homework again(贪心)
- HDU1789Doing Homework again(贪心+sort)
- HDU1789:Doing Homework again
- HDU1789--Doing Homework again
- Doing Homework again
- Doing Homework again(dp)
- hdu1789 Doing Homework again
- HDU Doing Homework again
- hdu Doing Homework again
- Doing Homework again
- 【1789 Doing Homework again】
- 编译原理与编译构造 课堂笔记9
- Windows消息机制
- LeetCode 2 Add Two Number
- java--charAt()
- META-INF文件夹是干啥的,META-INF文件夹的作用, META-INF文件夹能删吗
- HDU1789Doing Homework again
- PS(三)
- vue.js初学3之vue指令②
- OSPF笔记-8
- hadoop入门八(搭建环境)
- LaTeX公式语法1
- 在linux安装依赖于python包时,报错libpython.a: conld not read symbols
- java链表结构基本形式(学习笔记)
- 某些电脑下从注册表中获取的APPdata路径为空(来源于boost问题反馈)