题目链接：
http://acm.split.hdu.edu.cn/showproblem.php?pid=3661
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
Hint:
题意：
有两个长度为N（N<=1000）的序列A和B，把两个序列中的共2N个数分为N组，使得每组中的两个数分别来自A和B，每组的分数等于max(0,组内两数之和-t)，问所有组的分数之和的最小值
题解：
贪心。
代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1000+10;
int a[maxn],b[maxn];
int main()
{
int n,t;
while(scanf("%d%d", &n,&t)!=EOF)
{
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);
int sum=0;
for(int i=0,j=n-1;i<n;i++,j--)
{
if(a[i]+b[j]>t)
sum+=(a[i]+b[j]-t);
else
sum+=0;
}
printf("%d\n",sum);
}
}