Assignments |
| Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 85 Accepted Submission(s): 65 |
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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.
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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).
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Output
For each test case output the minimum Overtime wages by an integer in one line.
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Sample Input
2 5 |
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Sample Output
4 |
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Source
2010 Asia Regional Harbin
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Recommend
lcy
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/*
题意：有n个工作a，需要xi时间来完成，n个工作b，需要yi时间来完成，每天要有工人
来做工作a，b如果a，b的时间超过T，那么就要支付工人超出时长的额外金币，问最少
支付工人多少额外的金币。
初步思路：贪心一下，a从小到大排序，b从大到小排序，然后贪
*/
#include<bits/stdc++.h>
#define N 1100
using namespace std;
bool cmp1(int a,int b){
return a>b;
}
bool cmp2(int a,int b){
return a<b;
}
int n,t;
int a[N],b[N];
int main(){
freopen("in.txt","r",stdin);
while(scanf("%d%d",&n,&t)!=EOF){
for(int i=;i<n;i++)
scanf("%d",&a[i]);
for(int i=;i<n;i++)
scanf("%d",&b[i]);
sort(a,a+n,cmp1);
sort(b,b+n,cmp2);
int cur=;
for(int i=;i<n;i++){
if(a[i]+b[i]>t){
cur+=a[i]+b[i]-t;
}
}
printf("%d\n",cur);
}
return ;
}