区间调度问题

问题描述：

给定一组区间，每个区间都有一个开始时间和结束时间，目标是选择尽可能多的互不重叠的区间

区间调度实际上是在解决一组区间中，哪些区间可以同时选择而不起冲突的问题。

解决方法：

**贪心策略：**选择结束时间最早且与已选区间不重叠的区间。

例题--【CSES】Movie Festival

题目描述

In a movie festival n movies will be shown. You know the starting and ending time of each movie. What is the maximum number of movies you can watch entirely?

输入

The first input line has an integer n(1 ≤ n ≤ 2*105): the number of movies.

After this, there are n lines that describe the movies. Each line has two integers a and b(1 ≤ a < b ≤ 109): the starting and ending times of a movie.

输出

Print one integer: the maximum number of movies.

样例输入

3

3 5

4 9

5 8

样例输出

2

代码

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
const ll N=200010;
struct movie{
	int a,b;
}m[N];
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n;
	cin>>n;
	for (int i=1;i<=n;i++){
		cin>>m[i].a>>m[i].b;
	}
	sort(m+1,m+n+1,[](const movie &x,const movie &y){
		if (x.b==y.b)
		return x.a>y.a;
		return x.b<y.b;
	});
	int ans=0,end=0;
	for (int i=1;i<=n;i++){
		if (m[i].a>=end){
			ans++;
			end=m[i].b;
		}
	}
	cout<<ans;
}