UVA 11389 - The Bus Driver Problem

 

IIU C   ONLINE   C ON TEST  2 008

Problem E: The Bus Driver Problem

Input: standard input

Output: standard output

 

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 exceedsd, 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 aren space separated integers which are the lengths of the morning routes given in meters. Similarly the third line hasn 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

Output for Sample Input

2 20 5

10 15

10 15

2 20 5

10 10

10 10

0 0 0

50

0

 

Problem setter: Mohammad Mahmudur Rahman

题意：n个司机，ｎ条下午路线和ｎ条夜晚路线，给每个司机安排一条下午和夜晚路线，使得刚好每条路线都有一个司机，每个司机的总时间如果超过了d，则需给司机超出部分每小时ｒ元，最优安排使得总支付费用最少。输入n d r 和2*ｎ条线路的时间

思路：一看这么简单，还以为错的，结果真的这么简单。。

直接贪心，平均分配。一短一长的搭配

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define FOR(i, s, t) for(int (i)=(s); (i)<(t); (i)++)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const double PI = (4.0*atan(1.0));const int maxn = 100 + 20;int A[maxn];int B[maxn];int main() {    int n, d, r;    while(scanf("%d%d%d", &n, &d, &r) != EOF && !(!n && !d && !r)) {        for(int i=0; i<n; i++) scanf("%d", &A[i]);        for(int i=0; i<n; i++) scanf("%d", &B[i]);        sort(A, A+n);        sort(B, B+n);        int ans = 0;        for(int i=0; i<n; i++) {            int t = A[i] + B[n-i-1];            if(t > d) ans += (t-d) * r;        }        printf("%d\n", ans);    }    return 0;}