zoukankan      html  css  js  c++  java
  • HDOJ 1789 Doing Homework again

    按惩罚由大到小排序,惩罚高的先完成,完成的时间尽量推后。

    Doing Homework again

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4276    Accepted Submission(s): 2497


    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
     
    Author
    lcy
     
    Source
     
    Recommend
    lcy
     
     1 #include <iostream>
     2 #include <cstring>
     3 #include <algorithm>
     4 
     5 using namespace std;
     6 
     7 struct k
     8 {
     9     int t;
    10     int p;
    11 }hw[1111];
    12 
    13 bool cmp(k a,k b)
    14 {
    15 if(a.p!=b.p)
    16     return a.p>b.p;
    17 else
    18     return a.t<b.t;
    19 }
    20 
    21 int main()
    22 {
    23     int T;
    24     cin>>T;
    25 while(T--)
    26 {
    27     memset(hw,0,sizeof(hw));
    28     int n;
    29     cin>>n;
    30     for(int i=0;i<n;i++)
    31         cin>>hw[i].t;
    32     for(int i=0;i<n;i++)
    33         cin>>hw[i].p;
    34 
    35     sort(hw,hw+n,cmp);
    36 
    37     int tp[10000];
    38     memset(tp,0,sizeof(tp));
    39 
    40     for(int i=0;i<n;i++)
    41     {
    42         for(int j=hw[i].t;j>=1;j--)
    43         {
    44             if(tp[j]==0)
    45                 {
    46                     tp[j]=1;
    47                     hw[i].p=0;
    48                     break;
    49                 }
    50         }
    51     }
    52 
    53     int ans=0;
    54     for(int i=0;i<n;i++)
    55         ans+=hw[i].p;
    56 
    57     cout<<ans<<endl;
    58 
    59 }
    60 
    61     return 0;
    62 }
  • 相关阅读:
    LinQ表达式的一点点总结(二)select中新建对象
    给自己的博客添加分享到功能
    职场日记2上班第一天
    清楚屏幕右侧变化的数据Application.Current.Host.Settings.EnableFrameRateCounter = true;
    基于委托的异步
    C#中的装箱与拆箱
    关于java的初始化问题
    StreamReader类以及其方法ReadLine,Read,ReadToEnd的分析
    WP7的控件开发入门(二)
    单元测试的阶段性总结
  • 原文地址:https://www.cnblogs.com/CKboss/p/3113661.html
Copyright © 2011-2022 走看看