HDU 5126 stars cdq分治+树状数组

题目链接：点击打开链接

题意：

T个case

n个操作

1、 (x,y,z) 在三维平面的点上增加1

2、询问区间范围内的权值和。

思路：

cdq分治套cdq分治，然后套树状数组即可。。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <vector>
#include <string>
#include <time.h>
#include <math.h>
#include <iomanip>
#include <queue>
#include <stack>
#include <set>
#include <map>
const int inf = 1e8;
const double eps = 1e-8;
const double pi = acos(-1.0);
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != '-' && (c<'0' || c>'9')) c = getchar();
	sgn = (c == '-') ? -1 : 1;
	ret = (c == '-') ? 0 : (c - '0');
	while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x <0) { putchar('-'); x = -x; }
	if (x>9) pt(x / 10);
	putchar(x % 10 + '0');
}
using namespace std;
const int N = 50010 << 3;
int maxn, c[N];
int lowbit(int x){ return x&-x; }
void add(int x,int val){
	while (x <= maxn)c[x] += val, x += lowbit(x);
}
int sum(int x){
	int ans = 0;
	while (x)ans += c[x], x -= lowbit(x);
	return ans;
}
typedef long long ll;
typedef pair<int, int> pii;
class Node{
public:
	int op, x, y, z, id, add, ans;
	Node(){}
	Node(int _op, int _x, int _y, int _z, int _id, int _add) : op(_op), x(_x), y(_y), z(_z), id(_id), add(_add), ans(0){}
};
int ans[N];
bool cmp(const Node &e1, const Node & e2) {
	if (e1.x == e2.x)return e1.id < e2.id;
	return e1.x < e2.x;
}
bool cmp2(const Node &e1, const Node & e2) {
	if (e1.y == e2.y)return e1.id < e2.id;
	return e1.y < e2.y;
}
Node q[N], t[N], v[N];
int n, top, tt;
void cal(int l, int r){ //x有序
	if (l == r)return;
	int mid = (l + r) >> 1;
	tt = 1;
	for (int i = l; i <= mid; i++)if (t[i].op == 1)v[++tt] = t[i];
	for (int i = mid + 1; i <= r; i++)if(t[i].op == 2)v[++tt] = t[i];	
	sort(v + 1, v + tt + 1, cmp2);
	for (int i = 1; i <= tt; i++)
	if (v[i].op == 1)
		add(v[i].z, 1);
	else
		ans[v[i].id] += sum(v[i].z) * v[i].add;
	for (int i = 1; i <= tt; i++)if (v[i].op == 1)add(v[i].z, -1);
	cal(l, mid); cal(mid + 1, r);
}
void solve(int l, int r){
	if (l == r)return;
	int mid = (l + r) >> 1;
	top = 1;
	for (int i = l; i <= mid; i++)if (q[i].op == 1)t[++top] = q[i];
	for (int i = mid + 1; i <= r; i++)if (q[i].op == 2)	t[++top] = q[i];	
	sort(t + 1, t + top + 1, cmp);
	cal(1, top);
	solve(l, mid); solve(mid + 1, r);
}
vector<int>G;
int main(){
	int T; rd(T);
	while (T-- > 0){
		rd(n);
		G.clear();
		int siz = 0;
		for (int i = 1, op, x1, x2, y1, y2, z1, z2; i <= n; i++)
		{
			rd(op);
			if (op == 1)
			{
				q[++siz].op = 1;
				ans[i] = -1;
				rd(q[siz].x); rd(q[siz].y); rd(q[siz].z); G.push_back(q[siz].z);
			}
			else {
				rd(x1); rd(y1); rd(z1); rd(x2); rd(y2); rd(z2);
				G.push_back(z2); G.push_back(z1); G.push_back(z1 - 1);
				ans[i] = 0;
				q[++siz] = Node(2, x2, y2, z2, i, 1);
				q[++siz] = Node(2, x1 - 1, y2, z2, i, -1);
				q[++siz] = Node(2, x2, y1 - 1, z2, i, -1);
				q[++siz] = Node(2, x1 - 1, y1 - 1, z2, i, 1);

				q[++siz] = Node(2, x2, y2, z1 - 1, i, -1);
				q[++siz] = Node(2, x1 - 1, y2, z1 - 1, i, 1);
				q[++siz] = Node(2, x2, y1 - 1, z1 - 1, i, 1);
				q[++siz] = Node(2, x1 - 1, y1 - 1, z1 - 1, i, -1);
			}
		}
		sort(G.begin(), G.end());
		G.erase(unique(G.begin(), G.end()), G.end());
		maxn = G.size();
		for (int i = 1; i <= siz; i++)q[i].z = lower_bound(G.begin(), G.end(), q[i].z) - G.begin() + 1;
		solve(1, siz);
		for (int i = 1; i <= n; i++)if (ans[i] >= 0)pt(ans[i]), putchar('\n');
	}
	return 0;
}