CDQ分治或树套树可以切掉
CDQ框架：

• 先分
• 计算左边对右边的贡献
• 再和

所以这个题可以一维排序，二维CDQ，三维树状数组统计
CDQ代码

# include <stdio.h>

# include <stdlib.h>

# include <iostream>

# include <algorithm>

# include <string.h>

# define IL inline

# define RG register

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

IL ll Read(){

    RG char c = getchar(); RG ll x = 0, z = 1;

    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;

    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + c - '0';

    return x * z;

}

const int MAXN(100010), MAXK(200010);

int n, k, t[MAXK], m;

struct Point{

    int a, b, c, cnt, id, ans;

    IL bool operator ==(RG Point B) const{  return a == B.a && b == B.b && c == B.c;  }

} p[MAXN], q[MAXN];

IL bool Cmp1(RG Point x, RG Point y){  return x.a < y.a || (x.a == y.a && (x.b < y.b || (x.b == y.b && x.c < y.c)));  }

IL bool Cmp2(RG Point x, RG Point y){  return x.b < y.b || (x.b == y.b && (x.c < y.c || (x.c == y.c && x.a < y.a)));  }

IL void Add(RG int x, RG int d){  for(; x <= k; x += x & -x) t[x] += d;  }

IL int Query(RG int x){  RG int cnt = 0; for(; x; x -= x & -x) cnt += t[x]; return cnt;  }

IL void CDQ(RG int l, RG int r){

    if(l == r) return;

    RG int mid = (l + r) >> 1; CDQ(l, mid); CDQ(mid + 1, r);

    sort(p + l, p + r + 1, Cmp2);//可以归并排序

    for(RG int i = l; i <= r; i++)

        if(p[i].id <= mid) Add(p[i].c, p[i].cnt);

        else p[i].ans += Query(p[i].c);

    for(RG int i = l; i <= r; i++)

        if(p[i].id <= mid) Add(p[i].c, -p[i].cnt);

}

int main(RG int argc, RG char* argv[]){

    n = Read(); k = Read();

    for(RG int i = 1; i <= n; i++) q[i] = (Point){Read(), Read(), Read()};

    sort(q + 1, q + n + 1, Cmp1);

    for(RG int i = 1; i <= n; i++)

        if(q[i] == p[m]) p[m].cnt++;

        else p[++m] = q[i], p[m].cnt = 1, p[m].id = m;

    CDQ(1, m);

    sort(p + 1, p + m + 1, Cmp1);

    for(RG int i = 1; i <= m; i++) t[p[i].ans + p[i].cnt - 1] += p[i].cnt;

    for(RG int i = 0; i < n; i++) printf("%d\n", t[i]);

    return 0;

}

附上树套树

# include <stdio.h>

# include <stdlib.h>

# include <iostream>

# include <string.h>

# include <algorithm>

# define IL inline

# define RG register

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

IL ll Read(){

    RG char c = getchar(); RG ll x = 0, z = 1;

    for(; c > '9' || c < '0'; c = getchar()) z = c == '-' ? -1 : 1;;

    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + c - '0';

    return x * z;

}

const int MAXN(100010);

int n, k, ans[MAXN], NOW;

struct YYB{

    int a, b, c;

    IL bool operator <(RG YYB B) const{

        if(a != B.a) return a < B.a;

        if(b != B.b) return b < B.b;

        return c < B.c;

    }

    IL bool operator ==(RG YYB B) const{

        return a == B.a && b == B.b && c == B.c;

    }

} yyb[MAXN];

struct SBT{

    SBT *ch[2], *fa;

    int size, val, cnt;

    IL SBT(RG SBT *f, RG int v, RG int d){  cnt = size = d; val = v; fa = f;  }

} *rt[MAXN << 1];

IL int Son(RG SBT *x){  return x -> fa -> ch[1] == x;  }

IL int Size(RG SBT *x){  return x == NULL ? 0 : x -> size;  }

IL void Updata(RG SBT *x){  x -> size = Size(x -> ch[0]) + Size(x -> ch[1]) + x -> cnt;  }

IL void Rot(RG SBT *x){

    RG int c = !Son(x); RG SBT *y = x -> fa;

    y -> ch[!c] = x -> ch[c];

    if(x -> ch[c] != NULL) x -> ch[c] -> fa = y;

    x -> fa = y -> fa;

    if(y -> fa != NULL) y -> fa -> ch[Son(y)] = x;

    x -> ch[c] = y; y -> fa = x;

    Updata(y);

}

IL void Splay(RG SBT *x, RG SBT *f){

    while(x -> fa != f){

        RG SBT *y = x -> fa;

        if(y -> fa != f)

            Son(x) ^ Son(y) ? Rot(x) : Rot(y);

        Rot(x);

    }

    Updata(x); rt[NOW] = x;

}

IL void Insert(RG SBT *x, RG int v, RG int d){

    if(rt[NOW] == NULL){  rt[NOW] = new SBT(NULL, v, d); return;  }

    while(2333){

        RG int id = v > x -> val;

        if(v == x -> val){  x -> size += d; x -> cnt += d; break;  }

        if(x -> ch[id] == NULL){

            x -> ch[id] = new SBT(x, v, d);

            x = x -> ch[id];

            break;

        }

        x = x -> ch[id];

    }

    Splay(x, NULL);

}

IL int Calc(RG SBT *x, RG int v){

    RG int cnt = 0;

    while(x != NULL){

        if(v >= x -> val) cnt += Size(x -> ch[0]) + x -> cnt;

        if(v == x -> val) break;

        x = x -> ch[v > x -> val];

    }

    return cnt;

}

IL void Modify(RG int id, RG int d){

    for(NOW = yyb[id].b; NOW <= k; NOW += NOW & -NOW) Insert(rt[NOW], yyb[id].c, d);

}

IL int Query(RG int id){

    RG int cnt = 0;

    for(RG int i = yyb[id].b; i; i -= i & -i) cnt += Calc(rt[i], yyb[id].c);

    return cnt;

}

int main(){

    n = Read(); k = Read();

    for(RG int i = 1; i <= n; i++)

        yyb[i] = (YYB){Read(), Read(), Read()};

    sort(yyb + 1, yyb + n + 1);

    for(RG int i = 1, sum = 0; i <= n; i++){

        sum++;

        if(yyb[i] == yyb[i + 1]) continue;

        ans[Query(i) + sum - 1] += sum;

        Modify(i, sum);

        sum = 0;

    }

    for(RG int i = 0; i < n; i++) printf("%d\n", ans[i]);

    return 0;

}