三维偏序

题目链接

曼波~

\(\Large \texttt{注意到直接对第一维排序可以去掉一维。}\)
\(\Large \texttt{注意到使用归并排序可以去掉第二维。}\)
\(\Large \texttt{注意到在归并排序的过程中对以第二维为横坐标，第三维为纵坐标二维数点就可以得到答案。}\)
代码：

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
struct node{
	int x, y, z, ans, cnt;
}a[N], b[N], tp[N]; 
int n, k, s[N], out[N], tot;
bool cmp(node a, node b){
	if(a.x == b.x && a.y == b.y) return a.z < b.z;
	return a.x == b.x ? a.y < b.y : a.x < b.x;
}
void modify(int x, int k){
	for(; x < N; x += x & -x){
		s[x] += k;
	}
}
int query(int x){
	int ans = 0;
	for(; x; x -= x & -x) ans += s[x];
	return ans;
}
void merge(int l, int r){
	int mid = (l + r) >> 1;
	if(l == r) return;
	merge(l, mid);
	merge(mid + 1, r);
	vector<node> v;
	for(int i = l; i <= mid; i++){
		v.push_back(a[i]);
		modify(a[i].z, a[i].cnt);
	}
	for(int i = r; i >= mid + 1; i--){
		while(!v.empty() && v.back().y > a[i].y){
			modify(v.back().z, -v.back().cnt);
			v.pop_back();
		}
		a[i].ans += query(a[i].z);
	}
	while(!v.empty()) modify(v.back().z, -v.back().cnt), v.pop_back();
	int i = l, j = mid + 1, cnt = l;
	while(i <= mid && j <= r){
		if(a[i].y <= a[j].y) tp[cnt++] = a[i++];
		else tp[cnt++] = a[j++];
	}
	while(i <= mid) tp[cnt++] = a[i++];
	while(j <= r) tp[cnt++] = a[j++];
	for(int i = l; i <= r; i++) a[i] = tp[i];
}
int main(){
	cin >> n >> k;
	for(int i = 1; i <= n; i++){
		cin >> b[i].x >> b[i].y >> b[i].z;
	}
	sort(b + 1, b + n + 1, cmp);
	for(int i = 1, last = 0; i <= n; i++){
		if(b[i].x == b[i + 1].x && b[i].y == b[i + 1].y && b[i].z == b[i + 1].z) continue;
		b[i].cnt = i - last;
		last = i;
		a[++tot] = b[i];
	}
	merge(1, tot);
	for(int i = 1; i <= tot; i++){
		out[a[i].ans + a[i].cnt - 1] += a[i].cnt;
	}
	for(int i = 0; i < n; i++){
		cout << out[i] << '\n';
	}
	return 0;
}