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POJ 2376 Cleaning Shifts C++版

POJ题解 Slyar 2698浏览 1评论

文章作者:姜南(Slyar) 文章来源:Slyar Home (www.slyar.com) 转载请注明,谢谢合作。

贪心算法

Description

Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T.

Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval.

Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.

Input

* Line 1: Two space-separated integers: N and T

* Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.

Output

* Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.

Sample Input

3 10
1 7
3 6
6 10

Sample Output

2

Slyar:活动安排问题的进阶版,简化成数学问题就是给你长度为M的区间和N条线段的起点和终点,问你最少需要多少条线段就可以把区间覆盖(线段可以重叠)…

算法不难,按起点排序,起点相同按终点排,反正就是找符合条件的线段里覆盖区间最长的就好…不过陷阱N多,哥没注意一下WA了N-1次…排序后第一个起始点必须是1,有可能出现只有一条线段就完全覆盖的情况等等…还有,丫的数据太大不能用stream,必须用stdio…

// This problem has huge input data,
// use scanf() instead of cin to read data to avoid time limit exceed. 
#include 
#include 
#include 

using namespace std;

typedef struct
{
	int start;
	int end;
}cow;

// 按start升序,start相同则按end降序
bool cmp(const cow &a, const cow &b)
{
	if (a.start == b.start)
	{
		return a.end > b.end;
	}
	return a.start < b.start;
}

int main()
{
	int n, t;
	cow array[25010];

	scanf("%d%d", &n, &t);

	for (int i = 0; i < n; i++)
	{
		scanf("%d%d", &array[i].start, &array[i].end);
	}

	sort(array, array+n, cmp);

	// 第一头牛如果不从1开始则肯定无法覆盖全部
	if (array[0].start == 1)
	{
		// 只有一头牛而且它能覆盖全部
		if (array[0].end == t)
		{
			cout << "1" << endl;
		}
		else
		{
			int j = 0;
			int ans = 1;
			int start, end;
			bool find = false;

			// 找2个标记寻找符合start条件里end最大的牛
			start = end = array[0].end;

			// find标记是否可以完全覆盖
			while (!find)
			{
				// 每次在end以后的位置找
				for (int i = j + 1; i < n; i++)
				{
					// start要在前一个范围内,end尽可能要大
					if (array[i].start-1 <= start && array[i].end > end)
					{
						end = array[i].end;
						j = i;
					}
				}

				// 没找到符合条件的牛,肯定无法覆盖
				// 找到结尾则返回成功
				if (start == end)
				{
					break;
				}
				else if (end == t)
				{
					find = true;
				}

				// 牛数量++
				ans++;

				// 把start的位置更新
				start = end;
			}

			if (find)
			{
				cout << ans << endl;
			}
			else
			{
				cout << "-1" << endl;
			}
		}
	}
	else
	{
		cout << "-1" << endl;
	}
	return 0;
}

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  1. line 64: array[i].start-1 <= start 为什么不是array[i].start <= start?
    anoy8年前 (2012-05-19)回复