本题一般有两种解法，差分约束/贪心。分析思路不在赘述，有很多大佬博客有讲解。
这里只给出实现代码

算法1

(差分约束) 50ms+

C++ 代码

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
const int MAXN = 50000+10;
const int MAXM = 100000+10;
ll m,ans,head[MAXN],tot,a,b,c,dist[MAXN],vis[MAXN];
struct Edge{
    ll next,to,dat;
}edge[MAXM];
void add(ll x,ll y,ll c){
    edge[tot].to=y;
    edge[tot].dat=c;
    edge[tot].next=head[x];
    head[x]=tot++;
}
void spfa(ll st) {
    memset(dist,-0x3f,sizeof(dist));
    memset(vis,0,sizeof(vis));
    dist[st]=0;
    queue<ll> q;
    q.push(st);
    vis[st]=true;
    while(!q.empty())
    {   ll t=q.front();
        q.pop();
        vis[t]=false;
        for(ll i=head[t]; i!=-1; i=edge[i].next)
        {
            ll j=edge[i].to,w=edge[i].dat;
            if(dist[j]<dist[t]+w)
             {
               dist[j]=dist[t]+w;
                if(!vis[j])
                {   q.push(j);
                    vis[j]=true;

                }
              //  path[y]=x;
             }
        }
    }

}
int main(){
    memset(head,-1,sizeof(head));
    tot=0;
    scanf("%lld",&m);
    ans=0;
    while(m--){
        scanf("%lld%lld%lld",&a,&b,&c);
        b++;
        ans=max(a,ans),ans=max(b,ans);
        add(a,b,c);
        add(0,a,0),add(0,b,0);
    }
    for(ll i=1;i<=ans;i++)
    add(i,i-1,-1),  add(i - 1, i, 0);
    spfa(0);

cout<<dist[ans]<<endl;

    return 0;
}

算法2

(贪心+线段树) 30ms+

C++ 代码

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
typedef struct Node {
    int l, r;
    int sum;
    int lazy;
}Tree;
Tree tree[50010*4];
void pushdown(int p){
    if(tree[p].lazy)
    {   tree[p<<1].lazy+=tree[p].lazy;
        tree[p<<1|1].lazy+=tree[p].lazy;
        tree[p<<1].sum+=tree[p].lazy*(tree[p<<1].r-tree[p<<1].l+1);
        tree[p<<1|1].sum+=tree[p].lazy*(tree[p<<1|1].r-tree[p<<1|1].l+1);
        tree[p].lazy=0;
    }

}
void build(int p, int l, int r){
    int mid;
    tree[p].l=l;tree[p].r=r;
    tree[p].lazy=0;
    if(l==r){ tree[p].sum=0; return;}
    mid = (l+r)/2;
    build(p<<1,l,mid);
    build(p<<1|1,mid+1,r);
    tree[p].sum=tree[p<<1].sum+tree[p<<1|1].sum;

}
void change_point(int p, int x, int num)
{   int mid;
    if(tree[p].l==tree[p].r){ tree[p].sum=tree[p].sum+num; return ;   }
    mid = (tree[p].l+tree[p].r)/2;
    if(x<=mid) change_point(p<<1, x, num);
    else change_point(p<<1|1, x, num);
    tree[p].sum=tree[p<<1].sum+tree[p<<1|1].sum;
}
int ask_point(int p,int x)
{       int mid;
    if(tree[p].l==tree[p].r)
        {   return tree[p].sum;
        }
        if(tree[p].lazy) pushdown(p);
        mid=(tree[p].l+tree[p].r)>>1;
        if(x<=mid) return ask_point(p<<1,x);
        else return ask_point(p<<1|1,x);
}
void update(int p,int L, int R,int c)
{   int mid;
    if(L<=tree[p].l&&R>=tree[p].r)
   {    tree[p].lazy+=c;
        tree[p].sum+=c*(tree[p].r-tree[p].l+1);
        return;

   }
   pushdown(p);
    mid=(tree[p].l+tree[p].r)>>1;
    if(L<=mid) update(p<<1,L,R,c);
    if(R>mid)  update(p<<1|1,L,R,c);
        tree[p].sum=tree[p<<1].sum+tree[p<<1|1].sum;
}
int find(int p, int x, int y){
    int mid;
    if(x<=tree[p].l&&tree[p].r<=y) return tree[p].sum;
    pushdown(p);
    mid = (tree[p].l+tree[p].r)/2;
    if(y<=mid) return find(p<<1, x, y);
    if(x>mid) return find(p<<1|1,x,y);
    return find(p<<1, x, mid) + find(p<<1|1,mid+1,y);
}
struct no{
    int l,r,c;
    bool operator < (const no &other) const{
        return r<other.r;
    }
}node[50010];
int n,maxm;
int main(){
    scanf("%d",&n);
    maxm=0;
    for(int i=0;i<n;i++)
   { scanf("%d%d%d",&node[i].l,&node[i].r,&node[i].c);
       maxm=max(maxm,node[i].r);
   }
    //maxm为最大的r，即线段树右边边界。

    sort(node,node+n);  //每个区间按r从小到大排序

    build(1,1,maxm);    //建树
    int ans=0;
    for(int i=0;i<n;i++)
    {   int l=node[i].l,r=node[i].r,c=node[i].c;
        int sum=find(1,l,r);      //区间[l,r]的和
        int m=max(0,c-sum);       //m为还差多少满足要求
        while(m)
        {   if(ask_point(1,r))  //当前点已经为1
            {
                r--;
            }
            else {             //当前点不为1，则置为1，m--
                change_point(1,r,1);
                r--;
                m--;
                ans++;
            }

        }

    }
    printf("%d\n",find(1,1,maxm));

    return 0;
}