UVA 11389 - The Bus Driver Problem (贪心)_the bus driver problem uva

本文链接：https://blog.youkuaiyun.com/accelerator_/article/details/17435087

本文探讨了一种方法，通过合理分配公交司机的白天和夜晚行驶路线，以减少因工作时间过长而产生的加班费用。利用贪心算法，确保每位司机的总行驶距离不超过设定限制，从而最大化地节省成本。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

I I U C O N L I N E C O N T E S T 2 0 0 8
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 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	Output for Sample Input
2 20 5 10 15 10 15 2 20 5 10 10 10 10 0 0 0	50 0

题意：n辆BUS，有n条白天路线，n条夜晚路线，距离上限d，超过上限每小时加工资r。现在求怎么去分配使得给加班工资最少。

思路:贪心。每个BUS选一条最短白天路线和一条最长晚上路线即可。因为对于最小白天路线，选最大的夜晚路线使得d最大，而且如果这样选都会超过d，那么其他的路线都会超过，那么选这个是最合适的。

代码：

#include <stdio.h>
#include <algorithm>
using namespace std;
const int N = 105;

int n, d, r, af[N], ni[N];

bool cmp(int a, int b) {
    return a > b;
}

void init() {
    for (int i = 0; i < n; i ++)
	scanf("%d", &af[i]);
    for (int i = 0; i < n; i ++)
	scanf("%d", &ni[i]);
}

int solve() {
    int ans = 0;
    sort(af, af + n);
    sort(ni, ni + n, cmp);
    for (int i = 0; i < n; i ++) {
	if (af[i] + ni[i] > d) {
	    ans += (af[i] + ni[i] - d) * r;
	}
    }
    return ans;
}

int main() {
    while (~scanf("%d%d%d", &n, &d, &r) && n + d + r) {
	init();
	printf("%d\n", solve());
    }
    return 0;
}