2010acm区域赛(哈尔滨)--Assignments(贪心水题)

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本文介绍了一个关于工厂中员工任务分配的问题，目标是最小化因加班而产生的额外费用。通过将A类任务按所需时间从小到大排序，B类任务从大到小排序，实现了最优的配对策略。

Problem Description

In a factory, there are N workers to finish two types of tasks (A and B). Each type has N tasks. Each task of type A needs xi time to finish, and each task of type B needs yj time to finish, now, you, as the boss of the factory, need to make an assignment, which makes sure that every worker could get two tasks, one in type A and one in type B, and, what's more, every worker should have task to work with and every task has to be assigned. However, you need to pay extra money to workers who work over the standard working hours, according to the company's rule. The calculation method is described as follow: if someone’ working hour t is more than the standard working hour T, you should pay t-T to him. As a thrifty boss, you want know the minimum total of overtime pay.

Input

There are multiple test cases, in each test case there are 3 lines. First line there are two positive Integers, N (N<=1000) and T (T<=1000), indicating N workers, N task-A and N task-B, standard working hour T. Each of the next two lines has N positive Integers; the first line indicates the needed time for task A1, A2…An (Ai<=1000), and the second line is for B1, B2…Bn (Bi<=1000).

Output

For each test case output the minimum Overtime wages by an integer in one line.

Sample Input

2 5
4 2
3 5

Sample Output

题目大意:一个工厂有N个员工，有A，B两种类型的任务，A需要xi的时间完成，B需要yi的时间完成，每个员工必须有两项任务分别为A和B，每个员工的工作时间为T，如果超过，老板要支付t-T的钱给员工，因为老板欲省钱，所以求最小的支付。本题明显是贪心的水题，直接把A由小到大排序，B由大到小排序，这样加起来的结果必定最小。下面为代码:

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1001;
int a[maxn];
int b[maxn];
int cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int m,n,i,j,sum,total;
    while(cin>>m>>n)
    {
        sum = 0,total = 0;
        for(i=0;i<m;i++)
            cin>>a[i];
        for(j=0;j<m;j++)
            cin>>b[j];
        sort(a,a+m);
        sort(b,b+m,cmp);
        for(i=0;i<m;i++)
        {
            sum = a[i] + b[i];
            //cout<<sum<<endl;
            if(sum>n)
            {
                total += sum - n;
            }
        }
        cout<<total<<endl;
    }
}