1133 Non-overlapping segments (greedy)

There are N line segments on the X-axis, and each line segment has a start point S and an end point E. The maximum number of non-overlapping line segments that can be selected. (Note: The starting point or the ending point overlaps, and it is not counted as overlapping).

For example: [1 5][2 3][3 6], you can choose [2 3][3 6], these two line segments do not overlap each other.

Input

Line 1: 1 number N, number of line segments (2 <= N <= 10000)
Lines 2 - N + 1: 2 numbers per line, start and end of the line segment (-10^9 <= S, E <= 10^9)

Output

Output the maximum number of line segments that can be selected.

Input example

3
1 5
2 3
3 6

Output example

2

AC: code

#include<stack>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct edge{
	int st,en;
}ss[10005];
int cmp(edge x,edge y){
	if(x.en == y.en) return x.st<y.st;
	return x.in<y.in;
}
intmain()
{
	int t;
	cin>>t;
	for(int i = 0;i<t;i++)
	{
		scanf("%d %d",&ss[i].st,&ss[i].en);
	}sort(ss,ss+t,cmp);
	stack<edge>v;
	v.push(ss[0]);
	for(int i = 1;i<t;i++){
		edge e = v.top ();
		if(e.en>ss[i].st){
			continue;
		} else {
			v.push(ss[i]);
		}
	}
	printf("%d\n",v.size());
	return 0;
}