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IIUC ONLINE CONTEST 2008 / UVa 11389 The Bus Driver Problem (贪心)

2013-10-03 21:12 411 查看

11389 - The Bus Driver Problem

Time limit: 1.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2384

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

一增一减必然是最优的，可以通过交换两个元素的位置来反证。

完整代码：

/*0.016s*/

#include<cstdio>
#include<algorithm>
#include<functional>
using namespace std;

int a[105], b[105];

int main()
{
	int n, d, r, sum, temp;
	while (scanf("%d%d%d", &n, &d, &r), n)
	{
		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, greater<int>());
		sum = 0;
		for (int i = 0; i < n; ++i)
		{
			temp = a[i] + b[i] - d;
			if (temp > 0) sum += temp;
		}
		printf("%d\n", sum * r);
	}
	return 0;
}

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