P3810 【模板】三维偏序（陌上花开）cdq分治

传送门：https://www.luogu.org/problemnew/show/P3810

cdq分治的模板题，第一层外部排序，第二层cdq归并排序，这个时候不用考虑第一次的顺序，第三次用树状数组。

注意，不要用memset，用队列保存加上的值，最后在把加上的值减去就行了。

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 998244353;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}
#define MODmul(a, b) ((a * b >= mod) ? ((a * b) % mod + 2 * mod) : (a * b))
#define MODadd(a, b) ((a + b >= mod) ? ((a + b) % mod + 2 * mod) : (a + b))

/*-----------------------showtime----------------------*/
const int maxn = 1e5 + 9;
struct node {
    int x, y, z;
    int id;
} a[maxn], b[maxn], tmp[maxn];
bool cmp(node a, node b) {
    if (a.x != b.x) return a.x < b.x;
    if (a.y != b.y)
        return a.y < b.y;
    else
        return a.z < b.z;
}
int cnt[maxn], ans[maxn];

int sum[maxn * 2];
int lowbit(int x) {
    return x & (-x);
}
void add(int x, int c) {
    while (x < maxn * 2) {
        sum[x] += c;
        x += lowbit(x);
    }
}
int getsum(int x) {
    int res = 0;
    while (x > 0) {
        res += sum[x];
        x -= lowbit(x);
    }
    return res;
}
int lazy[maxn];
queue<int> que;
void cdq(int le, int ri) {
    int mid = (le + ri) >> 1;
    if (ri - le <= 0) return;
    cdq(le, mid);
    cdq(mid + 1, ri);

    //       memset(sum, 0, sizeof(sum));

    int p = le, q = mid + 1;
    int id = 0;
    while (p <= mid && q <= ri) {
        if (a[p].y <= a[q].y) {
            add(a[p].z, lazy[a[p].id]);
            que.push(p);
            tmp[++id] = a[p++];
        } else {
            cnt[a[q].id] += getsum(a[q].z);
            tmp[++id] = a[q++];
        }
    }
    while (p <= mid) tmp[++id] = a[p++];
    while (q <= ri) {
        cnt[a[q].id] += getsum(a[q].z);
        tmp[++id] = a[q++];
    }
    while (!que.empty()) {
        int u = que.front();
        que.pop();
        add(a[u].z, -lazy[a[u].id]);
    }

    for (int i = 1; i <= id; i++) a[i + le - 1] = tmp[i];
}
// int out[maxn];
map<p3, int> mp;

int main() {
    int n, k;
    scanf("%d%d", &n, &k);
    int tot = 0;
    rep(i, 1, n) {
        int x, y, z;
        scanf("%d%d%d", &x, &y, &z);
        //   read(x);read(y);read(z);
        p3 tp = p3(x, pii(y, z));
        if (mp.count(tp))
            lazy[mp[tp]]++, cnt[mp[tp]]++;
        else {
            tot++;
            a[tot].id = tot;
            a[tot].x = x;
            a[tot].y = y;
            a[tot].z = z;
            mp[tp] = tot;
            lazy[tot] = 1;
        }
    }
    sort(a + 1, a + 1 + tot, cmp);

    cdq(1, tot);

    rep(i, 1, tot) ans[cnt[a[i].id]] += lazy[a[i].id];
    rep(i, 0, n - 1) printf("%d\n", ans[i]);
    return 0;
}