三维偏序[cdq分治学习笔记]

三维偏序

就是让第一维有序 然后归并+树状数组求两维

cdq+cdq不会 告辞

#include <bits/stdc++.h>
// #define int long long
#define rep(a , b , c) for(int a = b ; a <= c ; ++ a)
#define Rep(a , b , c) for(int a = b ; a >= c ; -- a)
#define go(u) for(int i = G.head[u] , v = G.to[i] , w = G.dis[i] ; i ; v = G.to[i = G.nxt[i]] , w = G.dis[i])

using namespace std ;
using ll = long long ;
using pii = pair < int , int > ;
using vi = vector < int > ;

int read() {
  int x = 0 ; bool f = 1 ; char c = getchar() ;
  while(c < 48 || c > 57) { if(c == '-') f = 0 ; c = getchar() ; }
  while(c > 47 && c < 58) { x = (x << 1) + (x << 3) + (c & 15) ; c = getchar() ; }
  return f ? x : -x ;
}

template <class T> void print(T x , char c = '\n') {
  static char st[100] ; int stp = 0 ;
  if(! x) { putchar('0') ; }
  if(x < 0) { x = -x ; putchar('-') ; }
  while(x) { st[++ stp] = x % 10 ^ 48 ; x /= 10 ; }
  while(stp) { putchar(st[stp --]) ; } putchar(c) ;
}

template <class T> void cmax(T & x , T y) { x < y ? x = y : 0 ; }
template <class T> void cmin(T & x , T y) { x > y ? x = y : 0 ; }

const int _N = 1e6 + 10 ;
struct Group {
  int head[_N] , nxt[_N << 1] , to[_N] , dis[_N] , cnt = 1 ;
  Group () { memset(head , 0 , sizeof(head)) ; }
  void add(int u , int v , int w = 1) { nxt[++ cnt] = head[u] ; to[cnt] = v ; dis[cnt] = w ; head[u] = cnt ; }
} ;

const int N = 1e5 + 10  ;
typedef int arr[N] ;

int n , k ;
struct Node {
	int x , y , z , ans , cnt ;
	Node() {}
	bool operator < (const Node & other) const {
		if(x != other.x) return x < other.x ;
		if(y != other.y) return y < other.y ;
		return z < other.z ;
	}
	bool operator == (const Node & other) const {
		return (x == other.x) && (y == other.y) && (z == other.z) ;
	}
} a[N] , b[N] , tmp[N] ;
struct BIT {
	int c[N << 1] , mxk ;
	int low(int x) {
		return x & - x ;
	}
	void add(int x , int y) {
		if(! x) return ;
		for( ; x <= mxk ; x += low(x)) c[x] += y ;
	}
	int query(int x) {
		int ans = 0 ;
		for( ; x ; x ^= low(x)) ans += c[x] ;
		return ans ;
	}
} t ;
bool cmp(Node x , Node y) {
	if(x.y != y.y) return x.y < y.y ;
	if(x.z != y.z) return x.z < y.z ;
}
void solve(int l , int r){
//	printf("%d -> %d\n" , l , r) ;
	if(l == r) return ;
	int mid = l + r >> 1 ;
	solve(l , mid) ; solve(mid + 1 , r) ;
	int p = l ; rep(i , mid + 1 , r) {
		for( ; b[p].y <= b[i].y && p <= mid ; ++ p) t.add(b[p].z , b[p].cnt) ;
		b[i].ans += t.query(b[i].z) ;
	}
	rep(i , l , p - 1) t.add(b[i].z , -b[i].cnt) ;
	merge(b + l , b + mid + 1 , b + mid + 1 , b + r + 1 , tmp , cmp) ;
	rep(i , l , r) b[i] = tmp[i - l] ;
//	printf("%d -> %d\n" , l , r) ;
}
int ans[N] ;
signed main() {
	n = read() ; t.mxk = k = read() ;
	rep(i , 1 , n) { a[i].x = read() ; a[i].y = read() ; a[i].z = read() ; }
	sort(a + 1 , a + n + 1) ; rep(i , 1 , n) b[i] = a[i] ;
	int len = unique(b + 1 , b + n + 1) - b - 1 ; int tp = 1 ;
	rep(i , 1 , n) if(a[i] == b[tp]) ++ b[tp].cnt ; else ++ b[++ tp].cnt ;
	solve(1 , len) ;
	rep(i , 1 , len) { ans[b[i].ans + b[i].cnt - 1] += b[i].cnt ; }
	rep(i , 0 , n - 1) print(ans[i]) ;
	return 0 ;
}

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