The main idea of the topic ：

give n A mission , Every task has value , And it has to be in s[i] To t[i] At a certain time within the time , Ask the greatest value

Ideas ：

We first found out , At most n Time selected , So so many moments are useless
We discretize it directly
How to discretize ？
It must be discretized according to the first moment of each task .
And then we found out , Those of great value can be put forward . Because if a smaller one is put , It must be bigger and better .
Then we can do it step by step .

$code$

#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;

long long n;
struct node
{
    
	long long v, s, t;
}a[5010];
long long S[5010], b[10010];
long long f[5010];

bool cmp(node x, node y)
{
    
	return x.s<y.s;
}

bool cmp2(node x, node y)
{
    
	return x.v>y.v;
}

void get_point()
{
    
	for(long long i=1; i<=n; i++)
		S[i]=max(S[i-1]+1, a[i].s);
}

void lsh()
{
    
	for(long long i=1; i<=n; i++)
	{
    
		int j=1;
		while(S[j]<a[i].s) j++;
		a[i].s=j;
	}
}

bool find_(long long x, long long i)
{
    
	if(S[x]>a[i].t)
		return 0;
	if(f[x]==0)
	{
    
		f[x]=i;
		return 1;
	}
	long long j=f[x];
	if(a[i].t>a[j].t)
		return find_(x+1, i);
	else if(find_(x+1, j))
	{
    
		f[x]=i;
		return 1;
	}
	return 0;
}

int main()
{
    
	scanf("%lld", &n);
	for(long long i=1; i<=n; i++)
		scanf("%lld%lld%lld", &a[i].s, &a[i].t, &a[i].v);
	sort(a+1, a+1+n, cmp);
	get_point();
	sort(a+1, a+1+n, cmp2);
	lsh();
	long long ans=0;
	for(long long i=1; i<=n; i++)
		if(find_(a[i].s, i))
			ans+=a[i].v;
	printf("%lld", ans);
	return 0;
}