Assignments

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1637 Accepted Submission(s): 759


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

Assignments

將a、b兩種任務分別存在兩個數組中，然後分別排序，一個從小到大另一個從大到小排。再從1到就算完即可。因為我們可以假設當先取a中最小值時，將其與b中的最大值匹配一定不會虧，因為若與b中最大值匹配時間超過了T,這a中任何一個與之匹配都會超，若未超過，那自然最好（不用考慮若此時b中最大與a中次小匹配也不超的情況，因為這種情況下a中次小在a中最小與b中最大匹配完之後可以b和b中的次大完美匹配）。

代碼：

#include 
#include 
#define MAX 1100
using namespace std ;
bool cmp(const int a , const int b)
{
	return a>b ;
}
int a[MAX] , b[MAX] ;
int main()
{
	int n , t ;
	while(~scanf(%d%d,&n,&t))
	{
		for(int i = 0 ; i < n ; ++i)
		{
			scanf(%d,&a[i]) ;
		}
		for(int i = 0 ; i < n ; ++i)
		{
			scanf(%d,&b[i]) ;
		}
		sort(a,a+n);
		sort(b,b+n,cmp) ;
		int sum = 0 ;
		for(int i = 0 ; i < n ; ++i)
		{
			if(a[i]+b[i]>t)
			{
				sum += a[i]+b[i]-t ;
			}
		}
		printf(%d
,sum) ;
	}
	return 0 ;
}