【codeforces】Restaurant

最新推荐文章于 2020-07-26 23:29:10 发布

原创最新推荐文章于 2020-07-26 23:29:10 发布 · 242 阅读

0 ·

CC 4.0 BY-SA版权

编程语言同时被 2 个专栏收录

307 篇文章

订阅专栏

codeforces

76 篇文章

订阅专栏

本文介绍了一个算法问题，关于如何最大化接受餐厅的预订数量，在确保不同时段的预订不会发生时间上的冲突。通过一种有效的贪心算法策略，文章提供了解决方案，并附带了完整的C++代码实现。

Restaurant

Time Limit:4000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

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_ieach (1 ≤ l_i ≤ r_i ≤ 10⁹).

Output

Print the maximal number of orders that can be accepted.

Sample Input

Input

2
7 11
4 7

Output

Input

Output

Input

Output

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
#include<stdio.h>
#include<algorithm>
using namespace std;
struct stu
{
	int a,b;
}p[510000];
bool cmp(stu n,stu m)
{
	return n.b<m.b;
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=0;i<n;i++)
		scanf("%d%d",&p[i].a,&p[i].b);
		sort(p,p+n,cmp);
		int s=1,j=0;
		for(int i=1;i<n;i++)
		{
			if(p[i].a>p[j].b)
		{
			j=i;
			s++;
			}	
		}
	printf("%d\n",s);	
	}
	return 0;
}