#### The Bus Driver Problem UVA – 11389 | Solution

In a city there are n bus drivers. Also there are n morning bus routes and n afternoon bus routes with
various lengths. Each driver is assigned one morning route and 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 and 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

2 20 5
10 15
10 15
2 20 5
10 10
10 10
0 0 0

50
0

## Solution Code

``````#include<stdio.h>
void merge(int arr[], int l, int m, int r)
{
int i, j, k;
int n1 = m - l + 1;
int n2 =  r - m;
int L[n1], R[n2];
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];

i = 0;
j = 0;
k = l;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}

while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}

while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}

void mergeSort(int arr[], int l, int r)
{
if (l < r)
{

int m = l+(r-l)/2;

mergeSort(arr, l, m);
mergeSort(arr, m+1, r);

merge(arr, l, m, r);
}
}

void printArray(int A[], int size)
{
int i;
for (i=0; i < size; i++)
printf("%d ", A[i]);
printf("\n");
}

int main(){
int n,d,r,i,sum,temp;

while(scanf("%d%d%d",&n,&d,&r)&& n && d && r)
{
int m[n],e[n];
for(i=0;i<n;i++){scanf("%d",&m[i]);}
for(i=0;i<n;i++){scanf("%d",&e[i]);}
mergeSort(m, 0, n-1);
mergeSort(e, 0, n-1);
sum = 0;
for (i = 0; i < n; ++i)
{
temp = m[i] + e[n-i-1]-d;
if (temp > 0)
sum+=temp;
}
printf ("%d\n", sum * r);

}
return 0;

}``````
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