【UVa】11389 - The Bus Driver Problem

Problem here

Problem

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

Sample

input

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

output

50 
0

Solution

Sorting

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

bool cmp1(int &a, int &b){
    return a < b;
}
bool cmp2(int &a, int &b){
    return a > b;
}

int main(){

    int n, d, r;
    while(cin >> n >> d >> r){
        if(n == 0 && d == 0 && r== 0)
            break;

        vector<int> m, e;
        for(int i = 0; i < n; i++){
            int tm;
            cin >> tm ;
            m.push_back(tm);
        }
        for(int i = 0; i < n; i++){
            int te;
            cin >> te;
            e.push_back(te);
        }
        sort(m.begin(), m.end(), cmp1);
        sort(e.begin(), e.end(), cmp2);
        int ot = 0;
        for(int i = 0 ;i < n; ++i){
            int men = m[i] + e[i];
            if(men > d){
                ot += (men-d) * r;
            }
        }
        cout << ot << endl;
    }

    return 0;
}