UVA 11389 - The Bus Driver Problem (贪心)

最新推荐文章于 2020-03-04 23:27:10 发布

最新推荐文章于 2020-03-04 23:27:10 发布 · 257 阅读

本文探讨了如何在给定的公交车数量、白天和夜晚路线长度及加班费用的情况下，通过贪心算法为每名司机分配最短白天路线和最长夜晚路线，以最小化整体的加班费用。通过案例分析，展示了该方法的有效性和实现过程。

IIUCONLINE CONTEST2008
Problem E: The Bus Driver Problem
Input: standard input Output: standard output

In a city there arenbus drivers. Also there arenmorning bus routes &nafternoon 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 firstdhours at a flatrtaka / 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 integersn,dandr, as described above. In the second line, there arenspace separated integers which are the lengths of the morning routes given in meters. Similarly the third line hasnspace 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;
}