【LG4148】简单题

题面

洛谷

题解

\(kdt\)模板题呀。。。

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

inline int gi() {

	register int data = 0, w = 1;

	register char ch = 0;

	while (!isdigit(ch) && ch != '-') ch = getchar();

	if (ch == '-') w = -1, ch = getchar();

	while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();

	return w * data;

}

void chkmin(int &x, int y) { if (x > y) x = y; }

void chkmax(int &x, int y) { if (x < y) x = y; }

const int MAX_N = 4e5 + 5;

const double alpha = 0.75;

const int INF = 1e9;

struct Point { int x[2], w; } p[MAX_N];

struct Node {

	int mn[2], mx[2];

	int ls, rs, sum, size;

	Point tp;

} t[MAX_N];

int ans, N, M, WD, cur, top, rt, rub[MAX_N];

inline bool operator < (const Point &l, const Point &r) { return l.x[WD] < r.x[WD]; }

inline int newnode() {

	if (top) return rub[top--];

	else return ++cur;

}

void pushup(int o) {

	int ls = t[o].ls, rs = t[o].rs;

	for (int i = 0; i <= 1; i++) {

		t[o].mn[i] = t[o].mx[i] = t[o].tp.x[i];

		if (ls) chkmin(t[o].mn[i], t[ls].mn[i]), chkmax(t[o].mx[i], t[ls].mx[i]);

		if (rs) chkmin(t[o].mn[i], t[rs].mn[i]), chkmax(t[o].mx[i], t[rs].mx[i]);

	}

	t[o].sum = t[ls].sum + t[rs].sum + t[o].tp.w, t[o].size = t[ls].size + t[rs].size + 1;

}

int build(int l, int r, int wd) {

	if (l > r) return 0;

	int mid = (l + r) >> 1, o = newnode();

	WD = wd, nth_element(&p[l], &p[mid], &p[r + 1]), t[o].tp = p[mid];

	t[o].ls = build(l, mid - 1, wd ^ 1), t[o].rs = build(mid + 1, r, wd ^ 1);

	return pushup(o), o;

}

void pia(int o, int num) {

	if (t[o].ls) pia(t[o].ls, num);

	p[num + t[t[o].ls].size + 1] = t[o].tp, rub[++top] = o;

	if (t[o].rs) pia(t[o].rs, num + t[t[o].ls].size + 1);

}

void check(int &o, int wd) {

	if (alpha * t[o].size < t[t[o].ls].size || alpha * t[o].size < t[t[o].rs].size)

		pia(o, 0), o = build(1, t[o].size, wd);

}

void insert(int &o, int wd, Point tmp) {

	if (!o) return (void)(o = newnode(), t[o].ls = t[o].rs = 0, t[o].tp = tmp, pushup(o));

	if (tmp.x[wd] <= t[o].tp.x[wd]) insert(t[o].ls, wd ^ 1, tmp);

	else insert(t[o].rs, wd ^ 1, tmp);

	pushup(o), check(o, wd);

}

bool inner(int x1, int y1, int x2, int y2, int X1, int Y1, int X2, int Y2) { return (X1 >= x1 && X2 <= x2 && Y1 >= y1 && Y2 <= y2); }

int outter(int x1, int y1, int x2, int y2, int X1, int Y1, int X2, int Y2) { return (x1 > X2 || x2 < X1 || y1 > Y2 || y2 < Y1); }

int query(int o, int x1, int y1, int x2, int y2) {

	if (!o) return 0;

	int res = 0;

	if (inner(x1, y1, x2, y2, t[o].mn[0], t[o].mn[1], t[o].mx[0], t[o].mx[1])) return t[o].sum;

    if (outter(x1, y1, x2, y2, t[o].mn[0], t[o].mn[1], t[o].mx[0], t[o].mx[1])) return 0;

    if (inner(x1, y1, x2, y2, t[o].tp.x[0], t[o].tp.x[1], t[o].tp.x[0], t[o].tp.x[1])) res += t[o].tp.w;

    res += query(t[o].ls, x1, y1, x2, y2) + query(t[o].rs, x1, y1, x2, y2);

    return res;

}

int main () {

	N = gi();

	for ( ; ; ) {

		int op = gi(); if (op == 3) break;

		if (op == 1) insert(rt, 0, (Point){gi()^ ans, gi() ^ ans, gi() ^ ans});

		else {

			int x1 = gi() ^ ans, y1 = gi() ^ ans, x2 = gi() ^ ans, y2 = gi() ^ ans;

			printf("%d\n", ans = query(rt, x1, y1, x2, y2));

		}

	}

	return 0;

}