bzoj 1935 [Shoi2007]Tree 园丁的烦恼 cdq分治+树状数组

题面

题目传送门

解法

题目就是一个二维偏序问题

将某一个询问转化成二维前缀和的形式即可

还是用常数比较小的cdq分治吧，空间小，比较容易实现

坐标需要离散化

时间复杂度：\(O(n+m)\ log^2\ n\)

代码

#include <bits/stdc++.h>
#define N 2500010
using namespace std;
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Que {
	int x1, y1, x2, y2;
} q[N];
struct Node {
	int op, x, y, id, tim;
	bool operator < (const Node &a) const {
		return x < a.x;
	}
} a[N], t[N];
int n, m, sx, sy, x[N], y[N], ans[N], f[N], tx[N], ty[N];
int lowbit(int x) {return x & -x;}
void modify(int x, int v) {
	for (int i = x; i <= sy; i += lowbit(i))
		f[i] += v;
}
int query(int x) {
	int ret = 0;
	for (int i = x; i; i -= lowbit(i))
		ret += f[i];
	return ret;
}
void cdq(int l, int r) {
	if (l == r) return;
	int mid = (l + r) >> 1, tx = l, ty = mid + 1;
	for (int i = l; i <= r; i++)
		if (a[i].tim <= mid) t[tx++] = a[i];
			else t[ty++] = a[i];
	for (int i = l; i <= r; i++) a[i] = t[i];
	tx = l, ty = mid + 1;
	while (ty <= r) {
		while (tx <= mid && a[tx].x <= a[ty].x) {
			if (a[tx].op == 1) modify(a[tx].y, 1);
			tx++;
		}
		if (a[ty].op == 2) ans[a[ty].id] += query(a[ty].y);
		if (a[ty].op == 3) ans[a[ty].id] -= query(a[ty].y);
		ty++;
	}
	for (int i = l; i < tx; i++) if (a[i].op == 1) modify(a[i].y, -1);
	cdq(l, mid), cdq(mid + 1, r);
}
int main() {
	read(n), read(m);
	for (int i = 1; i <= n; i++) {
		read(x[i]), read(y[i]);
		tx[++sx] = x[i], ty[++sy] = y[i];
	}
	for (int i = 1; i <= m; i++) {
		read(q[i].x1), read(q[i].y1), read(q[i].x2), read(q[i].y2);
		tx[++sx] = q[i].x2, tx[++sx] = q[i].x1 - 1;
		ty[++sy] = q[i].y2, ty[++sy] = q[i].y1 - 1;
	}
	sort(tx + 1, tx + sx + 1); sort(ty + 1, ty + sy + 1);
	sx = unique(tx + 1, tx + sx + 1) - tx - 1;
	sy = unique(ty + 1, ty + sy + 1) - ty - 1;
	map <int, int> xx, yy;
	for (int i = 1; i <= sx; i++) xx[tx[i]] = i;
	for (int i = 1; i <= sy; i++) yy[ty[i]] = i;
	int len = 0;
	for (int i = 1; i <= n; i++)
		a[++len] = (Node) {1, xx[x[i]], yy[y[i]], 0, len};
	for (int i = 1; i <= m; i++) {
		int x1 = xx[q[i].x1 - 1], y1 = yy[q[i].y1 - 1];
		int x2 = xx[q[i].x2], y2 = yy[q[i].y2];
		a[++len] = (Node) {2, x2, y2, i, len}, a[++len] = (Node) {3, x1, y2, i, len};
		a[++len] = (Node) {3, x2, y1, i, len}, a[++len] = (Node) {2, x1, y1, i, len};
	}
	sort(a + 1, a + len + 1); cdq(1, len);
	for (int i = 1; i <= m; i++) cout << ans[i] << "\n";
	return 0;
}