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题解

处理区间问题，一般要先对区间按照右端点排序

因为这样排序之后，我如果顺序处理的话，我就算要在后面添加一些哨点，也和前面的区间已经没有关系了

接下来就是，对于每个区间右端点，我贪心地尽量在右边放哨点，这样为的是尽量对后边的区间提供便利

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 500010
#define maxm 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct SegmentTree
{
    ll sum[maxn<<2], L[maxn<<2], R[maxn<<2];
    pll mn[maxn<<2];
    void pushup(ll o)
    {
        mn[o]=min(mn[o<<1],mn[o<<1|1]);
        sum[o]=sum[o<<1]+sum[o<<1|1];
    }
    void build(ll o, ll l, ll r, ll* array=NULL)
    {
        ll mid(l+r>>1);
        L[o]=l, R[o]=r;
        if(l==r)
        {
            mn[o]=pll(0,-l);
            return;
        }
        build(o<<1,l,mid,array);
        build(o<<1|1,mid+1,r,array);
        pushup(o);
    }
    ll Sum(ll o, ll l, ll r)
    {
        ll mid(L[o]+R[o]>>1), ans(0);
        if(l<=L[o] and r>=R[o])return sum[o];
        if(l<=mid)ans+=Sum(o<<1,l,r);
        if(r>mid)ans+=Sum(o<<1|1,l,r);
        return ans;
    }
    pll Min(ll o, ll l, ll r)
    {
        ll mid(L[o]+R[o]>>1);
        pll ans(pll(linf,-1));
        if(l<=L[o] and r>=R[o])return mn[o];
        if(l<=mid)ans=min(ans,Min(o<<1,l,r));
        if(r>mid)ans=min(ans,Min(o<<1|1,l,r));
        return ans;
    }
    void chg(ll o, ll pos)
    {
        ll mid(L[o]+R[o]>>1);
        if(L[o]==R[o]){mn[o].first=1;sum[o]=1;return;}
        if(pos<=mid)chg(o<<1,pos);
        else chg(o<<1|1,pos);
        pushup(o);
    }
}segtree;
ll n, id[maxm], l[maxm], r[maxm], k[maxm], m;
int main()
{
    ll i, j;
    n=read(), m=read();
    rep(i,1,m)
    {
        l[i]=read(), r[i]=read(), k[i]=read();
        id[i]=i;
    }
    sort(id+1,id+m+1,[&](ll a, ll b){return r[a]<r[b];});
    segtree.build(1,1,n);
    rep(i,1,m)
    {
        // printf("l=%lld r=%lld k=%lld\n",l[id[i]],r[id[i]],k[id[i]]);
        while( segtree.Sum(1,l[id[i]],r[id[i]]) < k[id[i]] )
        {
            auto pr = segtree.Min(1,l[id[i]],r[id[i]]);
            // printf("(%lld,%lld)\n",pr.first,pr.second);
            segtree.chg(1,-pr.second);
        }
    }
    printf("%lld",segtree.Sum(1,1,n));
    return 0;
}