HDU 1789 Doing Homework again(馋)
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
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.
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.
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
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。
HDU 1789 Doing Homework again(馋)的更多相关文章
- hdu 1789 Doing HomeWork Again (贪心算法)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1789 /*Doing Homework again Time Limit: 1000/1000 MS ...
- HDU 1789 Doing Homework again (贪心)
Doing Homework again http://acm.hdu.edu.cn/showproblem.php?pid=1789 Problem Description Ignatius has ...
- HDU 1789 - Doing Homework again - [贪心+优先队列]
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1789 Time Limit: 1000/1000 MS (Java/Others) Memory Li ...
- HDU 1789 Doing Homework again(非常经典的贪心)
Doing Homework again Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Oth ...
- 题解报告:hdu 1789 Doing Homework again(贪心)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1789 Problem Description Ignatius has just come back ...
- HDU 1789 Doing Homework again(贪心)
Doing Homework again 这只是一道简单的贪心,但想不到的话,真的好难,我就想不到,最后还是看的题解 [题目链接]Doing Homework again [题目类型]贪心 & ...
- HDU 1789 Doing Homework again(贪心)
在我上一篇说到的,就是这个,贪心的做法,对比一下就能发现,另一个的扣分会累加而且最后一定是把所有的作业都做了,而这个扣分是一次性的,所以应该是舍弃扣分小的,所以结构体排序后,往前选择一个损失最小的方案 ...
- HDU 1789 Doing Homework again(排序,DP)
Doing Homework again Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Oth ...
- hdu 1789 Doing Homework again (Greedy)
Problem - 1789 继续贪心.经典贪心算法,如果数据比较大就要用线段树来维护了. 思路很简单,只要按照代价由大到小排序,然后靠后插入即可.RE了一次,是没想到deadline可以很大.如果d ...
随机推荐
- JS学习笔记 - 面向对象 - 选项卡 (普通选项卡改写)
选项卡3 <script> window.onload=function () { new TabSwitch('div1'); }; function TabSwitch(id) // ...
- Eclipse手动配置svn
1.在Eclipse根目录下建一个任意文件夹(如plugin),在该文件夹下建一个以该插件名命名的文件夹(如SVN).2.将下载下的插件文件解压,plugins和features文件夹复制到该文件夹下 ...
- 【Codeforces Round #445 (Div. 2) D】Restoration of string
[链接] 我是链接,点我呀:) [题意] 给你n个字符串. 让你构造一个字符串s. 使得这n个字符串. 每个字符串都是s的子串. 且都是出现次数最多的子串. 要求s的长度最短,且s的字典序最小. [题 ...
- maven 配置Project Facets时further configuration available不出来问题
如果下边的 further configuration available不出来 把Dynamic web module 去掉勾选,应用与项目,然后再点开项目的properties,再选中Dynami ...
- Docker基础(一)
1.安装:安装教程很多,Ubuntu14.04安装比较简单docker[之前使用Ubuntu13.04结果安装了好久也没有安装好,后来就直接是14,04了] 2.docker是容器,那么什么是容器? ...
- 杭电ACM1197——Specialized Four-Digit Numbers
题目的意思是从2992開始的四位数.每个四位数的10.12,16进制的数的每一位加起来都相等,就输出该数. 非常easy的一道题目. 以下的是AC的代码: #include <iostream& ...
- 用bootstrap做一个背景可轮转的登录界面
用bootstrap做一个背景可轮转的登录界面 一.总结 一句话总结:用css3的动画的 @keyframes 规则,制作轮转图. 1.用bootstrap做一个背景可轮转的登录界面? a.动画部分用 ...
- linux 安装完mysql 密码重置
If you have forgot the MySQL root password, can’t remember or want to break in….. you can reset them ...
- Mybatis全面详解——上(学习总结)
原文地址:https://blog.csdn.net/ITITII/article/details/79969447 一.什么是Mybatis 这里借用官网的一句话介绍什么是mybatis:MyBat ...
- C语言编程程序的内存怎样布局
在c语言中,每一个变量和函数有两个属性:数据类型和数据的存储类别. C语言中局部变量和全局变量变量的存储类别(static,extern,auto,register) 1. 从变量的作用域划分变量(即 ...