Restaurant （背包问题）

最新推荐文章于 2020-03-03 15:42:39 发布

杨.某某

最新推荐文章于 2020-03-03 15:42:39 发布

阅读量373

点赞数

CC 4.0 BY-SA版权

分类专栏：算法水题

本文链接：https://blog.youkuaiyun.com/qq_37823003/article/details/76400756

算法水题专栏收录该内容

55 篇文章

订阅专栏

本文探讨了一家餐厅如何在接受连续时间段预订的情况下，确定能够接受的最大预订数量，确保没有两个预订时间段重叠。通过排序和遍历算法实现了最优解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

点击题目打开链接

Restaurant

A restaurant received n orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the i-th order is characterized by two time values — the start time l_i and the finish time r_i (l_i ≤ r_i).

Restaurant management can accept and reject orders. What is the maximal number of orders the restaurant can accept?

No two accepted orders can intersect, i.e. they can't share even a moment of time. If one order ends in the moment other starts, they can't be accepted both.

Input

The first line contains integer number n (1 ≤ n ≤ 5·10⁵) — number of orders. The following n lines contain integer values l_i and r_i each (1 ≤ l_i ≤ r_i ≤ 10⁹).

Output

Print the maximal number of orders that can be accepted.

Example

Input

2
7 11
4 7

Output

Input

Output

Input

Output

题解：这是背包问题，直接做就可以

下面是我的代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
using namespace std;
struct P
{
	int star;
	int endd;
}num[900000];
bool cmp(P a,P b)
{
	return a.endd<b.endd;
}
int main()
{
	int n,k,p;
	while(scanf("%d",&n)!=EOF)
	{
		k=1;
		for(int i=0;i<n;i++)
		{
			scanf("%d%d",&num[i].star,&num[i].endd);
		}
		sort(num,num+n,cmp);
		p=num[0].endd;
		for(int i=1;i<n;i++)
		{
			if(num[i].star>p)
			{
				k++;
				p=num[i].endd;
			}
		}
		printf("%d\n",k);
	}
	return 0;
}