ACM: uva 11389 -&…

The Bus Driver Problem In a city there are n bus drivers. Also there are n morning bus routes & n afternoon bus routes with various lengths. Each driver is assigned one morning route & one evening route. For any driver, if his total route length for a day exceeds d, he has to be paid overtime for every hour after the first d hours at a flat r taka / hour. Your task is to assign one morning route & one evening route to each bus driver so that the total overtime amount that the authority has to pay is minimized.

Input

The first line of each test case has three integers n, d and r, as described above. In the second line, there are n space separated integers which are the lengths of the morning routes given in meters. Similarly the third line has n space separated integers denoting the evening route lengths. The lengths are positive integers less than or equal to 10000. The end of input is denoted by a case with three 0 s.

Output

For each test case, print the minimum possible overtime amount that the authority must pay.

Constraints

-           1 ≤ n ≤ 100 -           1 ≤ d ≤ 10000 -           1 ≤ r ≤ 5

 

sample input

2 20 5

10 15

10 15

2 20 5

10 10

10 10

0 0 0

 

sample outputs

50

0

 

题意: 给每位司机分配一个白天和晚上的行车路线, 每个路线时间给出, 并且如果司机工作时间为d,

     
超过时间d就算是加班工作, 每一小时就按照r工资计算, 现在要你安排使得最小的加班费用.

 

解题思路:

    
1. 问题很简单, 大小配即可.

 

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

#include <algorithm>

using namespace std;

#define MAX 105

int n, d, r;

int a[MAX], b[MAX];

bool cmp(int a, int b)

{

 return a > b;

}

inline int max(int a, int b)

{

 return a > b ? a : b;

}

int main()

{

// freopen("input.txt", "r", stdin);

 int i;

 while(scanf("%d %d %d", &n,
&d, &r) != EOF)

 {

  if(n == 0
&& d == 0
&& r == 0) break;

  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);

  

  int ans = 0;

  for(i = 0; i <
n; ++i)

   ans += max(0,
(a[i]+b[i]-d));

  printf("%d\n", ans*r);

 }

 return 0;

}