AssignmentsTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1075 Accepted Submission(s): 488 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
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//贪心题目，算是比较代码比较简单的了、刚开始看后我还真不会做、郁闷，其实就是尽可能把每个工人都把限定的时间用完了去，一个升序、一个降序、开始也不理解为什么这样就是对的了，貌似这样会让浪费尽可能小
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <cmath>
#define Y 1003
using namespace std;
bool cmp(const int&a,const int&b)
{
return a>b;
}
int main()
{
int a[Y],b[Y];
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);
for(sum=i=0;i<N;i++)
if(a[i]+b[i]>T)
sum+=a[i]+b[i]-T;
printf("%d\n",sum);
}
return 0;
}