Assignments
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 645 Accepted Submission(s): 300
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
Source
Recommend
lcy
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。
解法：贪心。将A B排序，A中最大的和B中最小的一组，A中第二大的和B中第二小的一组，以此类推。给出一个简单的证明：若有两组(a0,b0), (a1,b1)满足a0>=a1&&b0>=b1，这两组的得分为max(a0+b0-t,0)+max(a1+b1-t,0) >= max(a0+b1-t,0)+max(a1+b0-t,0)即(a0,b1),(a1,b0)的得分，所以交换b0 b1之后可以使解更优。
#include<stdio.h>
#include<algorithm>
using namespace std;
const int MAXN=1001;
int a[MAXN];
int b[MAXN];
bool cmp(int a,int b)
{
return a>b;
}
int main()
{
int n,t;
int sum;
while(scanf("%d%d",&n,&t)!=EOF)
{
int i;
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
scanf("%d",&b[i]);
sort(a,a+n);
sort(b,b+n,cmp);
sum=0;
for(i=0;i<n;i++)
{
if(a[i]+b[i]-t>0)
sum+=a[i]+b[i]-t;
}
printf("%d\n",sum);
}
return 0;
}