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最新推荐文章于 2022-03-21 16:33:19 发布

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分类专栏： hpu2016基础训练

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本文介绍了一个关于餐厅接受预订订单的问题，并提供了一种基于贪心算法的解决方案。该算法通过对比不同订单的时间段来确定餐厅可以接受的最大订单数。

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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

分析：贪心算法；

代码：

#include<cstdio>
#include<algorithm>
using namespace std;
struct time
{
	int st;
	int end_d;
}p[1000000];
bool cmp(time a,time b)
{
	return a.end_d <b.end_d ;
}
int main()
{int t,k=1;
scanf("%d",&t);
for(int i=0;i<t;i++)
	scanf("%d%d",&p[i].st ,&p[i].end_d );
	sort(p,p+t,cmp);
	int a=p[0].end_d ;
	for(int i=1;i<t;i++)
	if(p[i].st >a)
	{
		a=p[i].end_d ;
		k++;
	}
	printf("%d",k);
	return 0;
}