hdu 3661 Assignments

Assignments

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1191    Accepted Submission(s): 555

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

4

 

思路：贪心，a中最小的与b中最大的组合，即一个升序，一个降序排列即可。

因为假设(a0,b0),(a1,b1)两组数，a0>a1&&b0>b1则很明显max(a0+b0-t,0)+max(a1+b1-t,0)>=max(a0+b1-t,0)+max(a1+b0-t,0)。。

即(a1,b0),(a0,b1)比(a0,b0),(a1,b1)组合更好。

就上面的证明：假设a1+b1>=t; 则前面等于a0+b0+a1+b1-2t，后面也是。

            假设a1+b1<t,且其它都为正，前面等于a0+b0-t。后面a0+b0+a1+b1-2t==a0+b0-t+负数明显小于前者。

                        其它有为负数的就不用说了。。。

 

 

#include <iostream>
#include<algorithm>
using namespace std;
int cmp(int a,int b)
{
      return a>b;
}
int main()
{
      int i,n,t,sum;
      int a[1010],b[1010];
      while(cin>>n>>t)
      {
            sum=0;
           for(i=0;i<n;i++)
           cin>>a[i];
           for(i=0;i<n;i++)
           cin>>b[i];
           sort(a,a+n);
           sort(b,b+n,cmp);
           for(i=0;i<n;i++)
           {
                 if(a[i]+b[i]>t)
                 sum+=(a[i]+b[i]-t);
           }
           cout<<sum<<endl;
      }

}