hdu 3661 Assignments (贪心)

Assignments

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1463    Accepted Submission(s): 675

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

2010 Asia Regional Harbin

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  3665 3667 3664 3669 3668 

 

题意：给出n个工人，n个a类耗时和b类耗时，要求每个工人搭配一个a类的和b类的，每个工人、每个耗时都要有被分配。超出T的按超出的算。求超出最少的搭配。

贪心头加尾。不过总觉得有问题。

//203MS    236K    648 B    C++
#include<stdio.h>
#include<stdlib.h>
#define N 1005
int a[N];
int b[N];
int cmp(const void*a,const void*b)
{
    return *(int*)a-*(int*)b;
}
inline int max(int a,int b)
{
    return a>b?a:b;
}
int main(void)
{
    int n,t;
    while(scanf("%d%d",&n,&t)!=EOF)
    {
        for(int i=0;i<n;i++) 
            scanf("%d",&a[i]);
        for(int j=0;j<n;j++)
            scanf("%d",&b[j]);
        qsort(a,n,sizeof(a[0]),cmp);
        qsort(b,n,sizeof(b[0]),cmp);
        int vis[N]={0};
        int ans=0;
        for(int i=0;i<n;i++)
            ans+=max(0,a[i]+b[n-i-1]-t);
        printf("%d
",ans);
    }
    return 0;
}