洛谷 P3810 【模板】三维偏序（陌上花开）

原题链接
第一维直接排序，然后cdq分治+树状数组
对于分治的左右区间，区间内部按照第二维排序（已按第一维排序好了，就算打乱顺序，左右区间整体的第一维的偏序关系也不会受到影响）
然后遍历右区间的元素，把左区间的第二维小于当前元素的加入树状数组，统计答案即可，因为区间内部第二维都是单调不递减的，只需遍历一遍

#include<bits/stdc++.h>
using namespace std;
 
#define fr first
#define se second
#define et0 exit(0);
#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); i ++)
#define rrep(i, a, b) for(int i = (int)(a); i >= (int)(b); i --)
#define IO ios::sync_with_stdio(false),cin.tie(0);
 
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef unsigned long long ULL;
 
const int INF = 0X3f3f3f3f, N = 2e5 + 10, MOD = 1e9 + 7;
const double eps = 1e-7, pi = acos(-1);
 
int n, k;
 
struct NODE{
	int x, y, z, cnt, res;	
}a[N], b[N];
 
bool cmp1(NODE &A, NODE &B) {
	if (A.x == B.x) {
		if (A.y == B.y) return A.z < B.z;
		return A.y < B.y;
	}
	return A.x < B.x;
}
 
bool cmp2(NODE &A, NODE &B) {
	if (A.y == B.y) return A.z < B.z;
	return A.y < B.y;
}
 
int tr[N];
 
int lowbit(int x){
    return x & -x;
}
 
void insert(int x, int c){
    for(int i = x; i <= k; i += lowbit(i)) tr[i] += c;
}
 
int query(int x){
    int res = 0;
    for(int i = x; i >= 1; i -= lowbit(i)) res += tr[i];
    return res;
}
 
void cdq(int l, int r) {
	if (l == r) return;
	int mid = l + r >> 1;
	cdq(l, mid), cdq(mid + 1, r);
	
	sort(b + l, b + mid + 1, cmp2);
	sort(b + mid + 1, b + r + 1, cmp2);
	
	int tt = l;
	rep (i, mid + 1, r) {
		while (tt <= mid && b[i].y >= b[tt].y) {
			insert(b[tt].z, b[tt].cnt);
			tt++;
		}
		b[i].res += query(b[i].z);
	}
	rep (i, l, tt - 1) insert(b[i].z, -b[i].cnt);
}
 
void work() {
	cin >> n >> k;
	rep (i, 1, n) cin >> a[i].x >> a[i].y >> a[i].z;
	sort(a + 1, a + 1 + n, cmp1);
	int count = 0;
	rep (i, 1, n) {
		if (a[i].x != b[count].x || a[i].y != b[count].y || a[i].z != b[count].z) {
			count++;
			b[count].x = a[i].x, b[count].y = a[i].y, b[count].z = a[i].z;
			b[count].cnt = 1;
		}
		else b[count].cnt++;
	}
	cdq(1, count);
	vector<int> ans(n + 1);
	rep (i, 1, count) ans[b[i].res + b[i].cnt - 1] += b[i].cnt;
	rep (i, 0, n - 1) cout << ans[i] << endl;
}
 
signed main() {
	IO
 
	int test = 1;
 
	while (test--) {
		work();
	}
 
	return 0;
}