zoukankan      html  css  js  c++  java
  • Doing Homework again 贪心

    Doing Homework again

    题目描述

     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.

    输入

     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.

    输出

     For each test case, you should output the smallest total reduced score, one line per test case.

    示例输入

    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

    示例输出

    0
    3
    5

    提示

    #include<iostream>
    using namespace std;
    struct homework{
        int deadline, reduce;
    }a[1000], b;
    void quick_sort(homework s[], int l, int r){
        if(l < r){
            int i=l, j=r, x=s[l].deadline, y=s[l].reduce;
            b = s[l];
            while(i < j){
                while(i < j && (s[j].deadline > x || (s[j].deadline==x && s[j].reduce<y ) ) ) j--;
                if(i < j)  s[i++] = s[j];
                while(i < j && (s[i].deadline < x || (s[j].deadline==x && s[j].reduce>y )  ) ) i++;
                if(i < j)  s[j--] = s[i];
            }
            s[i] = b;
            quick_sort(s, l, i-1);
            quick_sort(s, i+1, r);
        }
    }
    int main(){
        int t;
        while(cin>>t) {
            while(t--){
                int n, i, Date = 1, score=0;
                cin>>n;
                for(i=0; i<n; i++)
                    cin>>a[i].deadline;
                for(i=0; i<n; i++)
                    cin>>a[i].reduce;
                quick_sort(a, 0, n-1);
                for(i=0; i<n; i++){
                    if(a[i].deadline >= Date) {
                        Date++;
                        continue;
                    }
                    int decline = a[i].reduce;
                    for(int j=0; j<i; j++)
                        if(a[j].reduce < decline )
                            decline = a[j].reduce;
                    score += decline;
                }
                cout<<score<<endl;
            }
        }
        return 0;
    }


  • 相关阅读:
    scala 基础
    Feign拦截器和解码器
    SpringCloud对使用者透明的数据同步组件
    POI SXSSF API 导出1000万数据示例
    HDFS文件浏览页返回上级目录功能
    Spring Security OAuth2.0
    Spring Security实现OAuth2.0授权服务
    Spring Security实现OAuth2.0授权服务
    Zookeeper学习笔记:简单注册中心
    Eclipse集成Git做团队开发:分支管理
  • 原文地址:https://www.cnblogs.com/Genesis2018/p/8304807.html
Copyright © 2011-2022 走看看